Physics 1st Paper Chapter 10 Past Paper Suggestions (PDF)

Physics 1st Paper Chapter-10 Ideal Gas and Kinetics of Gases  Important Topics of this Chapter for Creative Question of (c) & (d): Times Questions Board & Year the Questions Have Importance Topic...

Physics 1st Paper Chapter-10 Ideal Gas and Kinetics of Gases  Important Topics of this Chapter for Creative Question of (c) & (d): Times Questions Board & Year the Questions Have Importance Topic Appeared Appeared c d CQ T-01: Boyle’s & Charles’s RB’23, 21, 19; SB’23, 21; JB’23, 22 ;  Law and Law of 7 14 Ctg.B’22, 21, 19; DB’21; BB’21; CB’21, Pressure 19 CB’23, 21 ; SB’22; Din.B’22, 19;  T-02: Ideal Gass Equation 4 3 Ctg.B’19 ; JB’19 T-03: Root Mean Square Din.B’23, 22; MB’23; SB’22  2 1 Velocity T-04: Degrees of Freedom, DB’23; JB’23, 22, 19; CB’23, 19; Law of Equipartition Din.B’23 ; MB’23; RB’19  7 1 of Energy & Kinetic Energy  T-05: Mean Free Path - 1 Din.B’17 DB’23, 22, 21 ; Ctg.B’23, 22, 21; SB’23;  T-06: Relative Humidity 19 20 BB’23, 22; JB’23, 22, 21; RB’22, 21 ; CB’22, 21; MB’22, 21; Din.B’21, 19 T-01: Boyle’s & Charles’s Law and Law of Pressure 01. A helium gas balloon was filled at the surface of a saltwater ocean at a temperature of 27°C. One day, when the balloon was submerged under water to a temperature of 10°C, its diameter decreased to half of its value at the ocean surface. Again, when the balloon was submerged under water in a lake at the same depth, its diameter at that depth decreased to three-fourths of its value at the surface. Boltzmann constant is 1.38 × 10.23 JK −1 and atmospheric pressure is 105 Pa. Assume that the temperature of the lake is uniform. [RB’23] (c) Determine the volume of the balloon at a depth of 5 m in the lake. 3 (d) At a depth of 5 m, will the volume of the balloon be the same in both the ocean and the lake? Analyze mathematically. 4 Solution (c) Pressure at the surface, P1 = 105 Pa; Initial volume= V1; Density of the water of the lake, ρ = 1000 kgm−3, let h = 5 m, Pressure at the bottom, P2 = P1 + hρg = 105 + 5 × 1000 × 9.8 = 1.49 × 105 Pa P If final volume is V2 , then P1 V1 = P2 V2 ⇒ V2 = P1 V1 = 0.67V1 (Ans.) 2 1 (d) Assuming density of water in the lake, ρ = 1000 kg m−3, pressure at depth h, P2 = P1 + hρg = 105 + 9800h 3× r At the bottom, diameter = d′; radius = r′ 3d r′ According to question, d′ = ⇒ = 2 volume = V2 4 2 4 ′ 3r 4 ′3 4 ∴ r = ; V2 = πr = π × r ∴ V2 = V1 27 3 27 At the surface, diameter = d; radius = r 4 3 3 64 64 volume = V1 64 64 Now, P1 V1 = P2 V′ ⇒ P = P2 ⇒ × 105 = 105 + 9800h ⇒ h = 13. 98 ≈ 14 m 27 1 27 4 In case of the ocean, P1 = 105 Pa; T1 = 27°C = 300 K; V1 = πr 3 3 4 r 3 1 P2 = 105 + hρg = 105 + 14 × 9.8 × ρ = 105 + 137.2ρ; V2 = 3 π (2) = 8 V1 ; T2 = 10°C = 283 K V P1 V1 P2 V2 105 ×V2 (105 +137.2×ρ 1 ) Now, T1 = T2 ⇒ 300 = 283 8 ⇒ ρ = 4771.62 kg m−3 Let us assume, the initial volume on the ocean surface is V1 and P1 = 105 Pa Pressure at a depth of 5 m, P2 = P1 + hρg = 105 + 5 × 4771.62 × 9.8 = 3. 33 × 105 Pa P If the final volume is V2 , then P1 V1 = P2 V2 ⇒ V2 = P1 V1 = 0.3V1 [Assuming the temperature remains 2 constant] 27 From ‘c’ we get, Volume of the balloon at a depth of 5 m, V ′ = 64 V1 Volume of the balloon 5m beneath the ocean 0.3V1 Then, Volume of the balloon 5m beneath the lake = 27 = 0.71 ≠ 1 V 64 1 ∴ The volume will not be same in both cases. 02. At a certain location, a diver can explore underwater up to 5 m deep without an oxygen cylinder. A balloon was filled with air at air pressure of 105 Pa after which it had a volume of 10−3 m3. If the balloon is submerged there, the volume at a certain point is 4 × 10−4 m3. Density of water = 103 kgm−3 Acceleration due to gravity = 9.8 ms−2 [JB’23] (d) Can a diver perform underwater exploration without an oxygen cylinder where the volume of a balloon is 4 × 10−4 m3? Provide an opinion after mathematical analysis. 4 Solution (d) We know, P1 V1 = P2 V2 Pa = Pressure at the above surface = 105 Pa ⇒ Pa Va = (Pa + hρg)V2 Volume of the balloon at the above surface, ⇒ 105 × 10−3 = (105 + h × 1000 × 9.8) × 4 × 10−4 Va = 10−3 m3 ⇒ 105 + h × 1000 × 9.8 = 250000 Changed volume, V2 = 4 × 10−4 m3 ⇒ h = 15.31 m Depth, h =? ∴ h > 5m, So, the diver cannot perform without an oxygen cylinder. 03. Two students on a trip to an island observed the air bubbles in two ponds located side by side as follows. Atmospheric pressure = 105 Pa. [SB’23] Name of the pond Air bubble Density of water Freshwater pond When brought from the bottom of the water to the 1.0 × 103 kgm−3 surface, the volume becomes 4 times. Saltwater pond When brought from the bottom of the water to the 1.1 × 103 kgm−3 surface, the diameter becomes 2 times. (c) Determine the depth of the freshwater pond. 3 (d) Analyze the variation of water pressure at the bottom of the two ponds mathematically. 4 2  Solution (c) We know, P1 V1 = P2 V2 ⇒ Pa Va = PW VW Pressure on the upper surface of the pond Pa = ⇒ Pa Va = (Pa + hρg)Vw 105 Pa ⇒ 105 × 4VW = (105 + h × 103 × 9.8)VW Density of water, ρ = 1000 kgm−3 ∴ h = 30.61 m (Ans.) Depth, h =? (d) In case of 2nd pond: Pa Va = PW VW Pressure at the bottom of the lower pond, PW =? 5 3 ⇒ 10 × 2 × VW = PW × VW V ∝ d3 ⇒ 8 × 105 Pa = PW ∴ Va = 23 × VW From ‘c’, we get, depth of first pond= 30.61 m P1 = Pa + hρg = 105 + 30.61 × 103 × 9.8 = 399978 Pa ∆P = PW − P1 = 8 × 105 − 399978 = 400022 Pa (Ans.) 04. One day the air temperature above the water of Kaptai lake was 28°C, atmospheric pressure was 105 Pa and dew point was 10.5°C The temperature of water at the bottom of the lake was 7°C and the average density of water was 1000 kgm−3. On that day, an air bubble from the bottom of the lake reached to the surface of the water and its radius was tripled. [Saturated vapor pressure at 10°C, 11°C, and 28°C is 9.2, 9.9, and 28.5 mmHg respectively.] [Ctg.B’22] (d) Is it possible to determine the depth of the lake by taking the change in temperature into consideration? Analyze mathematically. 4 Solution (d) Let, depth of the lake = h Temperature of water at the bottom of the lake, T1 = (7 + 273) K = 280 K Volume of bubble at the bottom of the lake, V1 = V Pressure of water at the bottom of the lake, P1 = (105 + hρg) Pa Temperature of water at the surface of the lake, T2 = (28 + 273)K = 301 K Volume of bubble at the surface of the lake, V2 = (3)3 V1 = 27 V Atmospheric pressure of the lake, P2 = 105 Pa P1 V1 P2 V2 P2 V2 T 105 ×27V×280 We know, T1 = T2 ⇒ P1 = T2 × V1 ⇒ 105 + hρg = 301×V ⇒ hρg = 25.12 × 105 − 105 1 25.12×105 −105 ⇒h= 1000×9.8 ∴ h = 246.12 meter. Therefore, depth of the lake is 246.12 meter. That is, it is possible to determine the depth of the lake. 05. By taking a balloon filled with 200 gm of nitrogen to the bottom of a sea decreases the volume by half. The pressure at the surface of sea is 105 Nm−2 and temperature is 30°C. The temperature at the bottom of the sea is 15°C. Density of the water is 1000 kgm−3 ,g = 9.8 ms−2 , R = 8.314 Jmol−1 K −1. [JB’22] (d) Is it possible to determine the depth of the sea by considering the change in temperature? Explain mathematically. 4 3  Solution (d) P1 V1 P2 V2 105 ×V V (hρg+105 )× 2 P1 = pressure on the surface of sea = 105 Nm−2 We know, = ⇒ = T1 T2 303 288 V1 = volume on the surface of sea = V 105 ×2×288 T1 = temperature on the surface of sea = 303 K ⇒ hρg + 105 = 303 P2 = pressure at the bottom of sea ⇒ hρg = (1.90099 × 10 − 105 ) = 9.0099 × 104 5 9.0099×104 = (hρg + 105 )Nm−2 ∴h= = 9.19377 m V 1000×9.8 V2 = volume at the bottom of sea = 2 and ∴ It is possible to determine the depth of the sea. T2 = temperature at the bottom of sea = (273 + 15)K = 288 K 06. When an air bubble rises from the bottom of a lake to the surface of the water, its volume becomes double. Air pressure = 1.01 × 105 Pa. [DB’21] (c) Determine the depth of the lake. 3 (d) If the depth of the lake is 55 mm, what will be the change in volume of the air bubble having 2 cm radius when it rises from the bottom of the lake to the surface? Analyze mathematically. 4 Solution (c) Given that, V ′ = 2V, P = 1.01 × 105 + hρg, P′ = 1.01 × 105 Pa Now, PV = P′ V ′ ⇒ (1.01 × 105 + hρg)V = 2V(1.01 × 105 ) ⇒ 1.01 × 105 + 9800h = 2 × 1.01 × 105 ∴ h = 10.306 m (Ans.) (d) h = 55mm = 55 × 10−3 m, r1 = 2 cm = 0.02m 4 Now, P1 V1 = P2 V2 = πr13 × (1.01 × 105 + hρg) = P2 V2 3 4 ⇒ 3 π(0.02m)3 × (55 × 10−3 m × 9.8ms −2 × 1000 kgm−3 + 1.01 × 105 Pa) = V2 × 1.01 × 105 Pa ∴ V2 = 3.3689 × 10−5 m3 V1 = 3.351 × 10−5 m3 ∴ Change in volume = V2 − V1 = 1.79 × 10−7 m3 ; volume will increase. 07. A balloon filled with 5.1 liters of air at constant temperature was taken to a depth of 40 m under water, which caused its volume to become 1.1 liters. The maximum expansion capacity of the balloon is 9.5 liters and the acceleration due to gravity at that point is 9.8 ms −2. [RB’21] (c) What is the atmospheric pressure of that place according to the stem? 3 (d) If 1 liter of air is inserted into the above-mentioned balloon while it remains under water in a special arrangement and then released in sealed condition, will it reach the water surface intact? Analyze. 4 Solution (c) According to Boyle's law, at a certain temperature– P1 V1 = P2 V2 ⇒ (5.1 × 10−3 m3 ) × (P) = (1.1 × 10−3 m3 )(P + 40m × 1000 kgm−3 × 9.8 ms−2 ) 5.1 5.1 ⇒ 1.1 P = P + 392000 Pa ⇒ (1.1 − 1) P = 392000 Pa ∴ P = 107800 Pa ∴ Atmospheric pressure = 107800 Pa (𝐀𝐧𝐬. ) (d) From 'c', the pressure of the atmosphere, P = 107800Pa Now, P4 V4 = P3 V3 [∵ Temperature constant] ⇒ 107800Pa × V4 = (107800Pa + 392000Pa) × (2.1 × 10−3 m3 ) ⇒ V4 = 9.736364 × 10−3 m3 = 9.736364L ∴ V4 > 9.5 L ∴ The balloon will not come intact, it will burst. 4 08. A diver takes a gas-filled balloon to the bottom of a lake with 10m depth and commented that the volume of the balloon at the bottom of the lake is half that of the surface. Atmospheric pressure is 10−5 Nm−2 and density of lake water is 1000 kgm−3. [Ctg.B’21] (c) Determine the pressure acting on the balloon at the bottom of the lake. 3 (d) Verify the rationality of the comment of the diver with mathematical analysis. 4 Solution (c) ∴ Pressure at the bottom of the lake, P2 = P + hρg = 10−5 Nm−5 + 10m × 1000 kgm−3 × 9.8 ms−2 = 98000.00001 Nm−2 [Note: The question probably made an error of mentioning 10−5 Nm−2 instead of P = 105 Nm−2. In this case, P2 = (105 + 10 × 1000 × 9.8) Nm−2 = 198 × 103 Nm−2] (d) P1 = 10−5 Nm−2 , P2 = P + hρg Suppose, if the volume in the surface is V, then the If P = 105 Nm−2, If the volume at the bottom is volume at the bottom is V2 105 × V = (105 + 10 × 9.8 × 1000)V2 Now, PV = P2 V2 ⇒ V2 = 0.505V −5 (10−5 ⇒ 10 × (V) = + 10 × 9.8 × 1000) × V2 ∴ In this case the volume is almost half. The −10 ⇒ V2 = 1.02 × 10 V comment of the diver is rational. ∴ The volume of the balloon will be much smaller than half at the bottom. 09. One day, Kishore Pasha dived into a salt-water lake with 1020 kgm−3 density wearing a diver's suit. He noticed that the volume of the bubbles at the bottom of the lake became double when they reached the surface. Atmospheric pressure is 105 Pa, temperature of lake water is equal everywhere. [SB’21] (c) Determine the vertical distance traversed by the bubble. 3 (d) If the lake in the stem consisted of fresh water, then a diver would feel more comfortable in which type of lake? Analyze and give your opinion. 4 Solution (c) P1 V1 = P2 V2 ⇒ 2P1 V = P2 V ⇒ P2 = 2P1 ⇒ P1 + hρg = 2P1 Pressure at the surface, P1 = 105 Pa, Volume = 2V 105 Pressure at the bottom = P2 , Volume = V ⇒ hρg = 105 ⇒ h = 1020×9.8 ⇒ h = 10 m Density, ρ = 1020 kgm−3 ∴ Vertical distance = 10 m (Ans.) (d) Density of freshwater, ρ = 1000 kgm−3 Density of salt water, ρs = 1020 kgm−3 Pressure at ‘h’ depth underneath the fresh water, P = 105 + h × 1000 g Pressure at ‘h’ depth underneath the salt-water, Ps = 105 + h × 1020 g ∴ 1020 hg > 1000 h ∴ Ps > P That is, the pressure felt inside underneath salt-water is high but the pressure felt underneath fresh water is low. So, the diver will feel more comfortable underneath the fresh water lake. 5 10. A & B are two lakes where the volume increases by 4 times when a water bubble floats above from the bottom to the surface. The density of the water of the two lakes are 1000 kgm−3 & 1100 kgm−3. Atmospheric pressure is 105 Pa. [BB’21] (c) What is the pressure at the bottom of A lake? 3 (d) Among A & B, which lake has greater depth? Determine by mathematical explanation. 4 Solution (c) The pressure at the surface of A lake is P1 = 105 Pa, Volume V1 = 4V, Pressure at the bottom P2 , Volume 4V V2 = V; P1 V1 = P2 V2 ⇒ P2 = 105 × V ⇒ P2 = 4 × 105 Pa ∴ Pressure at the bottom = 4 × 105 Pa (Ans.) (d) For A lake, P1 + hρA g = 4P1 ⇒ hρA g = 3 × 105 ⇒ h = 30.6122 m 3×105 For B lake, P1 V1 = P2 V2 ⇒ P2 = 4P1 ⇒ P1 + hρB g = 4P1 ⇒ hρB g = 3 × 105 ⇒ hρB = 1100×9.8 = 27.83m ∴ A lake is deeper. 11. To determine the depth of a lake, a diver takes a balloon filled with 1 mole helium gas of 2.5 × 10−2 m3 volume at 1.013 × 105 Nm−2 pressure from the surface of the lake to the bottom, and as a result the volume decreases by half. Seeing this, the diver stated, “Depth of the lake is 9m.” Notable that the temperature of the lake and the gas is the same. [Density of the water of the lake = 1.1 × 103 kgm−3 , g = 9.8ms −2. R = 8.314 Jmol−1 K −1 , Atmospheric pressure on lake surface is 1.013 × 105 Nm−2 ] [CB’21] (d) Verify the statement of the diver. 4 Solution V (d) We Know that, P1 V1 = P2 V2 ⇒ 1.013 × 105 × V = P2 × 2 ⇒ P2 = 2 × 1.013 × 105 ⇒ P1 + hρg = 2 × 1.013 × 10 5 ρ = 1.1 × 103 kgm−3 1.013×105 ⇒ hρg = 2 × 1.013 × 105 − 1.013 × 105 ⇒ h = m = 1100 kgm−3 1100×9.8 ∴ h = 9.4 m P2 = P1 + hρg According to the statement of the diver, h = 9 m. ∴ So, the statement of the diver is not correct. 12. Temperature of water at the bottom and surface of a lake are 8°C and 30°C respectively. A balloon of volume 2 L is released from the bottom of the lake. Maximum expansion ability of the balloon is 15 L. Atmospheric pressure on the surface of the lake is 105 Nm−2 , depth of the lake is 15 m and density of water is 1000 kgm−3. [RB’19] (d) Find the possibility of bursting of the balloon after coming to the surface of the lake with mathematical explanation. 4 Solution P1 V1 P2 V2 (d) We know, T1 = T2 ; P1 = 105 ; T1 = 273 + 30 = 303 K ∴ P2 = (105 + 15 × 1000 × g); V2 = 2 L; T2 = 273 + 8 = 281 K P2 V2 T ∴ V1 = T2 × P1 = 5.32 L ∴ The balloon will not burst. 1 6 13. 24 gm of oxygen gas is filled in a balloon at 30°C temperature and 2 atm pressure. Mass of 1 mol of oxygen gas is 32 gm, and the atmospheric pressure at the surface of water of any pond is 1.5 atm. Density of water 1050 kgm−3 and depth 20 m. The atmospheric pressure over the surface of another pond is 1.2 atm, density of water 1000 kgm−3 and depth 25 m. [1 atm = 1.013 × 105 Pa, R = 8.314 Jmol−1 k −1 and g = 9.8 ms −2 ]. [Ctg.B’19] (d) In which pond of the stem, volume of gas in the balloon would be smaller at the bottom of water? Explain mathematically. 4 Solution P0 2×101325 (d) If the volume of the first pond is V1 then, V1 = P +h V0 = 1.5×101325+20×1050×9.8 V0 = 0.5664V0 1 1 ρ1 g P0 2×101325 If the volume of the second pond is V2 then, V2 = P +h V0 = 1.2×101325+25×1000×9.8 V0 = 0.5528V0 2 2 ρ2 g ∴ Second pond will have less volume. 14. An underwater diver can go up to 6m deep without an oxygen cylinder for exploring underwater. The volume of a balloon filled with gas at 105 Pa pressure at a place ‘A’ on the upper surface of water is 10−3 m3. If the balloon is submerged in water at place ‘B’, its volume becomes 5 × 10−4 m3. [Density of water = 103 kg m−3 , Acceleration due to gravity = 9.8 ms−2 ] [CB’19] (d) Can the diver keep exploring without an oxygen cylinder at place ‘B’? Give your opinion by showing mathematical analysis. 4 Solution P2 V1 10−3 (d) At B, P2 = P1 + hρg. Now, = = ∴ P2 = 2 × 105 P1 V2 5 × 10−4 P2 − P1 So, P2 = P1 + hρg ⇒ ρg = h ∴ h = 10.2 m ∴ Oxygen cylinder will be required. T-02: Ideal Gass Equation 01. A liter flask can contain 1.32 × 10−3 kg nitrogen gas at normal temperature and pressure. When the flask is half-filled with hot water of 90°C, the gas remains in thermal equilibrium with the water. [CB’23] (d) Verify the correctness of the statement that the flask will not contain half of the previous amount of gas after half-filling with hot water. 4 Solution w RT 1.32×10−3 8.314×273 (d) Previously, PV = nRT ⇒ V = = × = 1.056 × 10−3 m3 M P 28×10−3 101325 V 1.056×10−3 After half filling with water, Vnew = 2 = 2 = 5.28 × 10−4 m3 w′ Later, PVnew = nRT′ ⇒ 101325 × 5.28 × 10−4 = × 8.314 × (273 + 90) M w′ ⇒ 28 = 0.0177 ⇒ w ′ = 0.49 g = 4.96 × 10 kg −4 w′ 4.96×10−4 Now, = = 0.37 w 1.32×10−3 ∴ Will not have half gas of before. 7 02. A cylinder of volume 1 × 10−2 m3 is filled with Oxygen at a temperature of 300 K at 2.5 × 105 Nm−2 pressure. By keeping the temperature constant, the pressure was found to be 1.3 × 105 Nm−2 after some amount of oxygen was used. Molecular mass of oxygen is 32 gmol−1. [SB’22] (d) Is it possible to determine what amount of usage of oxygen gas is done from the description of the stem? Give mathematical analysis. 4 Solution (d) Volume of cylinder, V is constant Temperature, T and constant initial mole number of oxygen is, P1 V 2.5×105 ×1×10−2 n1 = RT = 8.314×300 = 1.0023254 mol If the mole number is changed to n2 then the changed pressure of gas is P2 , P2 V 1.3×105 ×1×10−2 ∴ n2 = RT = 8.314×300 = 0.5212092 mol ∴ Amount of used oxygen = n1 − n2 = (1.0023254 − 0.5212092)mol = 0.4811162 mol = 15.3957g 03. A container is filled with Nitrogen gas at 20 atm pressure at a temperature of 27°C. Afterwards gas of half of the mass is removed and the temperature of the rest of the gas is increased up to 87°C. [Din.B’22] (d) In the changed situation will the pressure of the gas be more or less than the pressure in the initial stage? Give your opinion with mathematical analysis. 4 Solution n (d) Because of removing half of the gas, Changed mole number of the gas = 2 Changed temperature, T ′ = (273 + 87)K = 360 K n P ×360 P 3 3 We know, P ∝ nT ∴ P2 = 2n×300 ⇒ P2 = 5 ∴ P2 = 5 P1 1 1 3 ∴ The pressure of the gas in the changed situation is times of the initial stage and P2 < P1. 5 04. To determine the depth of a lake, a diver takes a balloon filled with 1 mole helium gas of 2.5 × 10−2 m3 volume at 1.013 × 105 Nm−2 pressure from the surface of the lake to the bottom, and as a result the volume decreases by half. Seeing this, the diver stated, “Depth of the lake is 9m.” Notable that the temperature of the lake and the gas is the same. [Density of the water of the lake = 1.1 × 103 kgm−3 , g = 9.8ms −2. R = 8.314 Jmol−1 K −1 , Atmospheric pressure on lake surface is 1.013 × 105 Nm−2 ] [CB’21] (c) Determine the temperature of the gas. 3 Solution PV 1.013×105 ×2.5×10−2 (c) If the temperature of is T, PV = nRT ⇒ T = nR = 1×8.314 K = 304.61K ; T = 31.61°C (Ans.) 05. At 30°C temperature and 2 atm pressure, a balloon contains 24 gm of oxygen gas. The mass of 1 mole oxygen is 32 gm, on the other hand, the air pressure at the surface of a lake is 1.5 atm, density of water is 1050 kgm−3 and depth is 20m. In another lake, the air pressure at the surface is 1.2 atm, density of water is 1000kgm−3 and depth is 25 m. [Ctg.B’19] [1 atm = 1.013 × 105 Pa, R = 8.314 Jmol−1 K −1 and g = 9.8 ms−2 ] (c) Determine the volume of the gas in the stem. 3 Solution nRT 24 1 (c) PV = nRT ⇒ V = P = 32 × 0.0821 × 303 × 2 = 9.3286L 8 06. A cylinder of volume 0.6 m3 at room temperature is filled with 800 gm of methane (CH4 ) gas at a pressure of 202650 Pa at a definite location. The teacher told his students that rain is likely to happen only when the dew point of the place is 11.5°C and the relative humidity of the place is above 60%. The saturated water vapor pressure at 11°C, 12°C, 19°C and 20°C was found to be 9.84 mm(Hg), 10.52 mm(Hg), 16.46 mm(Hg) and 17.54 mm(Hg) respectively. The molecular mass of methane is 16 gmol−1. [Din.B’19] (c) Determine the room temperature at that location. 3 Solution w PVM 202650×0.6×16 (c) According to the ideal gas equation, PV = nRT ⇒ PV = RT ∴ T = = = 292.49K (𝐀𝐧𝐬. ) M wR 800×8.314 07. [JB’19] m1 = 30 g C = 575 ms-1 R = 8.31 m2 = 48 g TK Jmole-1K-1 TK Fig 1: Nitrogen gas Fig 2: Carbon di oxide (c) What will be the temperature of the gas kept in the cylinder? 3 Solution (c) For the gas in the cylinder, R = 8.31 Jmol−1 K −1 ; C = 575 ms−1 ; M = 28 × 10−3 kg 3RT C2 M 5752×28×10−3 We know, C = √ M ⇒ 3R =T⇒T= 3 × 8.31 = 371.34 K = 98.34°C (Ans.) T-03: Root Mean Square Velocity 01. A student in a Physics laboratory converted 4 kg of oxygen in a volume of 6.2 × 10−7 m3 at a pressure of 0.62 mHg and a temperature of 27°C to STP. At the end of the experiment the teacher commented that both the volume of the gas and the root mean square velocity decrease. [Molecular mass of oxygen is 32 kg and universal gas constant R = 8.31 JK −1 mol−1 ] [Din.B, MB’23] (d) Was the teacher's comment of the stem correct? Explain with mathematical analysis. 4 Solution P1 V1 P2 V2 0.62×6.2×10−7 0.76×V2 (d) T1 = T2 ⇒ 300 = 293 ⇒ V2 = 4.6 × 10−7 m3 Given that, initital pressure, P1 = 0.62 mmHg 3RT1 3×8⋅314×3Ω Again, Crms = √ M =√ 32×10−3 = 483.56m s−1 Final pressure, P2 = 0.76 mmHg Initial temperature, T1 = 27°C = 300 K 3RT2 3×8⋅314×273 Crms = √ =√ = 461.28 ms −1 Final temperature, T2 = 273 K M 32×10−3 ∴ Crms < Crms−1 এবং V2 < V1 ∴ The teacher’s comment of the stem was correct. 02. A cylinder of volume 1 × 10−2 m3 is filled with Oxygen at a temperature of 300 K at 2.5 × 105 Nm−2 pressure. By keeping the temperature constant, the pressure was found to be 1.3 × 105 Nm−2 after some amount of oxygen was used. Molecular mass of oxygen is 32 gmol−1. [SB’22] (c) Determine what was the root mean square velocity of the molecules of the oxygen gas mentioned in the stem before use. 3 9  Solution 3RT (c) Root mean square velocity of the molecules, crms = √ M 3×8.314×300 Temperature, T = 300 K Crms = √ 32×10−3 ms −1 = 483.561 ms−1 (Ans.) Molecular mass, M = 32g = 32 × 10−3 kg crms =? 03. A container is filled with Nitrogen gas at 20 atm pressure at a temperature of 27°C Afterwards gas of half of the mass is removed and the temperature of the rest of the gas is increased up to 87°C [Din.B’22] (c) Determine the root mean square velocity of the molecules of the nitrogen gas in the initial stage? 3 Solution (c) Given that, Pressure, P = 20 atm Temperature, T = 27 + 273 = 300 K M = 28g = 28 × 10−3 kg 3RT 3×8.314×300 Root mean square velocity, Crms = √ M =√ 28×10−3 = 516.95 ms −1 (Ans.) T-04: Degrees of Freedom, Law of Equipartition of Energy & Kinetic Energy 01. On a scorching hot summer day, Mimi and her classmates were conducting an experiment in the laboratory with 1000 grams of methane gas in a cylinder, where the molar mass of methane is 16 g/mol. At atmospheric temperature and pressure, the teacher informed the students that if the relative humidity exceeds 70% at any location, there is a possibility of rain. At that location, the dew point is 12.5°C and air temperature is 20.6°C. The vapor pressure in the atmosphere at the temperatures 12 °C, 13 °C, 20°C and 21°C are 10.85 mm Hg, 14.5 mm Hg, 18.5 mmHg and 20.55 mmHg respectively. [DB’23] (c) Determine the kinetic energy of the methane gas in the cylinder. 3 Solution (c) We know, f 6 w E = 2 nRT = 2 × M × RT = 3 × 1000 × 8.314 × 293.6 E = Kinetic energy 16 f = degrees of freedom of methane= 6 = 457.6857 × 103 J (Ans.) n = number of moles, T = 20. 6°C = 293.6 K 02. A liter flask can contain 1.32 × 10−3 kg nitrogen gas at normal temperature and pressure. When the flask is half-filled with hot water of 90°C, the gas remains in thermal equilibrium with the water. [CB’23] (c) Determine the kinetic energy of a nitrogen molecule at the temperature of water. 3 Solution We know, E = 2 kT = 2 × 1.38 × 10−23 × 363 Temperature of water, T = 90°C = 273 + 90 = 363K f 5 (c) Boltzmann constant, k = 1.38 × 10−23 Jkg −1 mol−1 = 1.25 × 10−20 J (Ans.) Degree of freedom, f = 5 10  03. At a certain location, a diver can explore underwater up to 5 m deep without an oxygen cylinder. A balloon was filled with air at air pressure of 105 Pa after which it had a volume of 10−3 m3. If the balloon is submerged there, the volume at a certain point is 4 × 10−4 m3. Density of water = 103 kgm−3 [JB’23] Acceleration due to gravity = 9.8 ms−2 (c) Find out the kinetic energy of the gas inside the balloon above the water surface. 3 Solution (c) We know, 3 E = 2 PV Pressure at the above surface, P = 105 Pa 3 Volume of the balloon, V = 10−3 m3 = 2 × 105 × 10−3 = 150 J (Ans.) Kinetic energy, E =? 04. In a Physics laboratory, a student transferred Oxygen of 6.2 × 10−7 m3 volume and 4 kg at 0.62 mHg pressure and 27°C temperature into STP. After experimentation, the teacher commented that volume of gas and its root mean square velocity both get reduced. [Molecular mass of Oxygen is 32 and Ideal Gas Constant R = 8.31 JK −1 mol−1 ] [Din.B, MB’23] (c) Determine the kinetic energy of Oxygen at initial stage. 3 Solution f (c) Kinetic Energy, F = 2 nRT Given that, 5 w Mass of used Oxygen, w = 4 gm ⇒ 2 × M RT Molecular mass of Oxygen, M = 32 g 5 4 = 2 × 32 × 8 ⋅ 314 × 300 = 779.4375 J (Ans.) Initial Temperature, T1 = 27°C = 300 K 05. By taking a balloon filled with 200 gm of nitrogen to the bottom of a sea decreases the volume by half. The pressure at the surface of sea is 105 Nm−2 and temperature is 30°(c) The temperature at the bottom of the sea is 15°C. Density of the water is 1000 kgm−3 , g = 9.8 ms −2 , R = 8.314 Jmol−1 K −1. [JB’22] (c) Determine the total kinetic energy of the nitrogen gas on the sea surface. 3 Solution 200 (c) Mole number of N2 gas, n = 28 = 7.143 mole 3 Total energy of N2 gas on the sea surface, ET = 2 nRT 5 T = 30°C = (30 + 273)K = 303 K ET = 2 × 7.143 × 8.314 × 303 = 4.5 × 104 J (Ans.) 06. Temperature of water at the bottom and surface of a lake are 8°C and 30°C respectively. A balloon of volume 2 L is released from the bottom of the lake. Maximum expansion ability of the balloon is 15 L. Atmospheric pressure on the surface of the lake is 105 Nm−2 , depth of the lake is 15 m and density of water is 1000 kgm−3. [RB’19] (c) Find the change in kinetic energy of the enclosed molecules in the balloon. 3 Solution PV (P+hρg)×V 5 5 5 (c) n = RT = RT = 0.21145 mol; Ek1 = 2 nRT1 ; Ek2 = 2 nRT2 ∴ ΔEk = 2 nRΔT = 96.7 J (Ans.) 11 07. [JB’19] m1 = 30 g C = 575 ms-1 R = 8.31 m2 = 48 g TK Jmole-1K-1 TK Fig 1: Nitrogen gas Fig 2: Carbon di oxide (d) Which container’s gas has greater kinetic energy? Explain mathematically. 4 Solution 3RT C2 M 5752×28×10−3 30 (d) We know, C = √ M ⇒ 3R =T⇒T= 3 × 8.31 = 371.34 K; For N2 , n = 28 = 1.07 mol 5 5 In case of Nitrogen, E = 2 nRT = 2 × 1.07 × 8.31 × 371.34 = 8265.63 J 48 Mole number of CO2 , n = = 1.0909 mol 44 In case of Carbon-dioxide, E = 3 nRT = 3 × 1.091 × 8.31 × 371.34 = 10099.097 J ∴ ECO2 > EN2 ; That is, the kinetic energy of second cylinder is higher. 08. An underwater diver can go up to 6m deep without an oxygen cylinder for exploring underwater. The volume of a balloon filled with gas at 105 Pa pressure at a place ‘A’ on the upper surface of water is 10−3 m3. If the balloon is submerged in water at place ‘B’, its volume becomes 5 × 10−4 m3. [Density of water = 103 kg m−2 , Acceleration due to gravity = 9.8 ms−2 ] [CB’19] (c) Determine the kinetic energy of the gas inside the balloon. 3 Solution (c) On the upper surface of water P = 105 Pa ; V = 10−3 m3 f 5 ∴ Ek = 2 PV = 2 × 105 × 10−3 = 250 J (Ans.) T-05: Mean Free Path 01. The volume of a gas cylinder is 1.5m3. Inside the cylinder 30 × 1025 molecules of a certain gas are confined at 27°C temperature. The radius of gas molecules is 25 × 10−10 m. Afterwards this gas filled cylinder is added to an empty cylinder of equal volume. [Din.B’17] (d) Will the mean free path of the gas be changed because of addition of an empty cylinder? Give your opinion based on mathematical analysis. Solution (d) Before attaching the empty cylinder, Volume of gas, V = 1.5m3 No. of molecules, N = 30 × 1025 ; Diameter of molecule, d = 25 × 10−10 m N 30×1025 ∴ Number of molecules in unit volume, n = = = 20 × 1025 V 1.5 ∴ According to Maxwell’s law, the mean free path of the gas molecules, 1 1 λ= = m = 1.8 × 10−10 m √2 π nd2 √2 π×20×1025 ×(25×10−10 )2 12  If empty cylinder of equal volume is attached, Volume of gas, V′ = 2V = 3m3 N 30×1025 ∴ Number of molecules in unit volume, n′ = V′ = 3 = 1026 ∴ According to Maxwell’s law, the mean free path of the gas molecules, 1 1 λ′ λ′ = = −10 )2 m = 3.6 × 10−10 m ∴ =2 √2rn d2 √2π×10 26 ×(25×10 λ Therefore, if empty cylinder is attached, then the mean free path of the gas molecules will be doubled. 4 T-06: Relative Humidity 01. On a scorching hot summer day, Mimi and her classmates were conducting an experiment in the laboratory with 1000 grams of methane gas in a cylinder, where the molar mass of methane is 16 g/mol. At atmospheric temperature and pressure, the teacher informed the students that if the relative humidity exceeds 70% at any location, there is a possibility of rain. At that location, the dew point is 12.5°C and air temperature is 20.6°C. The vapor pressure in the atmosphere at the temperatures 12 °C, 13 °C, 20°C and 21°C are 10.85 mm Hg, 14.5 mm Hg, 18.5 mmHg and 20.55 mmHg respectively. [DB’23] (d) In light of the stem determine whether there is a possibility of rain at that location, and give justification for your answer. 4 Solution (d) For a temperature change of (13 − 12) = 1°C, The increase in vapor pressure = (14.5 − 10.85) = 3.65 mm Hg ⸫ For a temperature change of (12.5 − 12) = 0.5°C, The increase in vapor pressure = 3.65 × 0.5 = 1.825 mm Hg Vapor pressure at dew point= 10. 85 + 1.825 = 12.675 mm Hg = f For a temperature change of (21 − 20) = 1°C, The increase in vapor pressure= (20.55 − 18.5) = 2.05 mm Hg ⸫ For a temperature change of (20.6 − 20) = 0.6°C, The increase in vapor pressure= 2.05 × 0.6 = 1.123 mm Hg ⸫ Vapor pressure at the temperature of the air= 18.5 + 1.123 = 19.73 mm Hg = F f 12.675 R = F × 100 = 19.73 × 100 = 64.24 % ∵ R < 70 % ⸫ There is no possibility of rain. 02. On any given day the temperature in Dhaka is 25°C and the dew point is 15.7°C. A hygrometer is placed in Kuakata on that same day, and it recorded a wet-bulb temperature of 20.1°C. Glacier’s constant at 25°C is 1.7. The vapor pressures at 15°C, 16°C, 20°C, 22°C and 25°C temperatures are 12.8 mmHgp, 13.63 mmHgP, 17.54 mmHg, 19.83 mmHgP and 23.8 mmHgP respectively. [Ctg.B’23] (c) Determine the relative humidity in Dhaka. 3 (d) On the day mentioned in the stem, would it be more comfortable in Dhaka or Kuakata? Analyze mathematically. 4 Solution (c) Given that, Vapor pressure at 15°C temperature is 12. 8 mmHgP Vapor pressure at 16°C temperature is 13.63 mmHgP 13  Now, For a change of (16 − 15)°C = 1°C, difference in pressure= (13.63 − 12.8) mmHgP = 0.83 mmHgP ∴ For a change of 0.3°C difference in pressure= 0.83 × 0.3 mmHgP = 0.249 mmHgP ∴ Vapor pressure at 15.7°C temperature = (13.63 − 0.249) mmHgP = 13.381 mmHgP Vapor pressure at dew point ∴Relative humidity of Dhaka, x = Vapor pressure at current temperature × 100% 13.381 = 23.8 × 100% = 56.22% (Ans.) (d) Dew point at Kuakata, θ = θ1 − G(θ1 − θ2 ) θ1 = 25°C = 25 − 1.7 × (25 − 20.1) θ2 = 20.1°C = 25 − 8.33 = 16.67°C G = 1.7 From ‘c; we get, For a change of 1°C change in pressure is 0.83 mmHgP For a change of 0.67°C change in pressure is= 0.83 × 0.67 mmHgP = 0.5561 mmHgP ∴ Vapor pressure at 16.67°C temperature, = (13.63 + 0.5561) mmHgP = 14.1861 mmHgP 14.1861 ∴ Relative humidity of Kuakata, x ′ = 23.8 × 100% = 59.61% From ‘c’ we get, Relative humidity of Dhaka, x = 56.22% x < x′ ∴ It will be more comfortable in Dhaka. 03. A meteorologist collected the following data through two wet and dry bulb hygrometers installed in Dhaka and Rajshahi on a day to prepare a daily report. [BB’23] Reading of Dry Bulb Reading of Wet Bulb Glaisher’s constant at air Location Thermometer Thermometer temperature Dhaka 28.6°C 20°C 1.664 Rajshahi 32.5°C 22°C 1.625 [At temperatures 14°C, 16°C, 28°C, 30°C, 32°C and 34°C saturated vapor pressure are 11.99, 13.63, 28.35, 31.83, 35.66 and 39.90 mm HgP respectively. Optimum humidity for human comfort ranges between 40% and 60%] (c) Determine the dew point of Dhaka that day. 3 (d) In the light of the stimulus, where will a person feel more comfortable? Analyze mathematically and give your opinion. 4 Solution (c) We know, θ1 − θ = G(θ1 − θ2 ) Here, ⇒ 28.6 − θ = 1.664(28.6 − 20) Dry bulb reading, θ1 = 28.6°C ⇒ θ = 28.6 − 14.3104 Wet bulb reading, θ2 = 20°C ⇒ θ = 14.2891°C (Ans.) Dew point, θ =? (d) From ‘c’, we get, dew point of Dhaka= 14.2891°C Increase of vapor pressure due to change of temperature by (30 − 28) = 2°C is = (31. 83 − 28.35) = 3.48 mmHg ⸫ Increase of vapor pressure due to change of temperature by (28.6 − 28) = 0.6 is 3.48 = 2 × 0.6 = 1.044 mmHg 14  Saturated vapor pressure of air, F = 28.35 + 1.044 = 29.394 mmHg Change of vapor pressure due to change of temperature by (16 − 14) = 2°C is = (13.63 − 11.99) = 1.64 mmHg ⸫Change of vapor pressure due to change of temperature by (14.2891 − 14) = 0.2891°C is 1.64 = × 0.2896 = 0.237472 mmHg 2 ⸫ Saturated vapor pressure at dew point, f = 12.227472 mm Hg f 12.227472×100 ⸫ R1 = F × 100 = 29.394 = 41.6% In Rajshahi: θ1 − θ = G(θ1 − θ2 ) ⇒ 32.5 − θ = 1.625 (32.5 − 22) ⇒ θ = 15.4375° C 39.90−35.66 Saturated vapor pressure at air temperature, F = 34−32 × (32.5 − 32) + 35.66 = 36.72 mmHg 13.63−11.99 Saturated vapor pressure at dew point, f = × (15.4375 − 14) + 11.99 = 13.16875 mmHg 16−14 f 13.16875 R 2 = F × 100 = 36.72 × 100 = 35.8630% Here, 40% < R1 < 60% and R 2 < 40% Therefore, a person will feel more comfortable in Dhaka. 04. The air temperature at a certain location is 20°C and the dew point is 7.4°C. The air temperature is lowered to 16°C. The vapor pressures of saturated water vapor at 7°C, 8°C and 20°C are 7.53 mm, 8.05 mm and 17.54 mm Hg, respectively. [JB’23] (c) What is the relative humidity of that area? 3 (d) Will the dew point change if the air temperature decreases while keeping the relative humidity constant? Analyze mathematically. 4 Solution (c) Increase of vapor pressure due to change of temperature by (8 − 7) = 1°C is = (8.05 − 7.53) = 0.52 mmHg Increase of vapor pressure due to change of temperature by (7.4 − 7) = 0.4°C is = (0.52 × 0.4) = 0.208 mmHg ∴ Saturated vapor pressure at dew point, f = 7.53 + 0.208 = 7.738 mm Hg Saturated vapor pressure at 20°C temperature, F = 17.54 mm Hg f 7.738 R = F × 100 = 17.54 × 100 = 44.12% (Ans.) (d) From ‘c’, Relative humidity = 44.12% Increase of vapor pressure due to change of temperature by (20 − 8) = 12°C is = (17.54 − 8.05)mm Hg = 9.49 mm Hg Increase of vapor pressure due to change of temperature by (20 − 16) = 4°C is 9.49 = × 4 = 3.16 mm Hg 12 ∴ Saturated vapor pressure at 16°C = 17.54 − 3.16 = 14.38 mm Hg f f Now, R = F × 100 ⇒ 44.12 = 14.38 × 100 = f = 6.344456 mm Hg For (8.05 − 7.53) = 0.52 mm Hg change, temperature decreases 1°C ∴ For (8.05 − 6.344456) = 1.705544 mm Hg change in saturated vapor pressure, temperature decreases 1 0.52 × 1.705504 = 3.28°C ∴ Dew point = (8 − 3.28) = 4.72°C ∴ Dew point will decrease. 15 05. In a recent experiment, the air temperature in Khagrachari was found to be 28°C and the dew point was 7.8℃. Suddenly, a cold wave caused the temperature to drop to 14℃. The saturated water vapor pressure at 7℃, 8℃ and 28°C is 7.51, 8.05, and 28.35 mm mercury respectively. [SB’23] (c) Calculate the relative humidity of Khagrachari. 3 (d) Will the dew point of the confined air in that place remain the same or different with the change of temperature? Give your opinion with mathematical reasoning. 4 Solution (c) Change of pressure due to temperature difference of (8 − 7)°C = 1°C is = (8.05 − 7.51) = 0.54 mm Hg 0.54 ∴ Change of pressure due to temperature difference of (7.8 − 7)°C = 0.8°C is = 1 × 0.8 = 0.432 mm Hg ∴Saturated vapor pressure at 7.8°C is, f = 7.51 + 0.432 = 7.942 mm Hg f R = × 100 = 7⋅942 × 100 = 28.014% (Ans.) F = Saturated vapor pressure at 28°C F 28⋅35 (d) The dew point of the confined air in that place will remain the same with the change of temperature. The temperature in that place has decreased, but the amount of water vapor has remained the same. We know that the temperature at which the air is saturated with water vapor is called the dew point. The amount of water vapor in that place is the same. As a result, the air in that place would be saturated with that water vapor at a temperature of 7.8°C. In other words, even if the temperature is brought to 14°C, the air would not be saturated with its water vapor. Only at 7.8°C will the air be saturated with its water vapor. However, the relative humidity will change. 28.35−8⋅05 Saturated vapor pressure at 14°C = 28−8 × 6 + 8.05 = 14.14 mm Hg From ‘c’ we get, at 7.8°C, f = 7.942 mmHg f 7⋅942 R = f × 100 = 14⋅14 × 100 = 56.16% m ∴ The dew point will remain the same at a temperature of 14°C because the amount of water vapor is the same. However, the relative humidity will not remain the same. 06. One day, the following information was obtained about a particular area: [DB’22] Inside the house: Dry bulb temperature = 32°C; Wet bulb temperature = 28°C Glaisher's constant at 32°C = 1.63 ; Saturated vapor pressure at 24°C = 22.38 mm Hg Saturated vapor pressure at 26°C = 25.21 mm Hg; Saturated vapor pressure at 32°C = 35.66 mm Hg. Temperature outside the house = 14°C; Saturated vapor pressure at 14°C = 12.0 mmHg Relative humidity = 80%. (c) Determine the dew point inside the house. 3 (d) Will wet clothes dry quickly outside or inside the house on that day? Verify mathematically. 4 Solution (c) We know, Glaisher’s equation related to dew point, θ = θ1 − G(θ1 − θ2 ) Glaisher’s constant, G = 1.63 Temperature of the dry bulb, θ1 = 32℃ Temperature of the wet bulb, θ2 = 28℃ Hence, θ = θ1 − G (θ1 − θ2 ) = 32° − 1.63(32° − 28°) ∴ Dew point, θ = 25.48℃ (Ans.) 16 (d) Dew point obtained from ‘c’, θ = 25.48℃ Now, Due to increase of (26 − 24)℃ = 2℃ temperature, change in vapor pressure = (25.21 − 22.38) mm Hg = 2.83 mm Hg. 2.83 Due to increase of 1℃ temperature, change in vapor pressure = 2 mm Hg ∴ Due to increase of (25.48 − 24)℃ = 1.48℃ temperature, 2.83 change in vapor pressure = 2 × 1.48 mm Hg = 2.0942 mm Hg. Vapor pressure at dew point, f = (22.38 + 2.0942) mm Hg = 24.4742 mm Hg Saturated vapor pressure at 32℃, F = 35.66 mm Hg f 24.4742 Relative humidity inside the house, R = F × 100% = 35.66 × 100% = 68.63% Relative humidity outside the house, R′ = 80% Relative humidity outside the house R′ > R; So, clothes will dry quickly inside the house. 07. On a particular day, the Meteorological Department of Rangpur observed that the reading obtained from the wet bulb and dry bulb of a hygrometer was 28°C and 26°C respectively. Glaisher’s factor at 28°C is 1.65. The saturated vapor pressure at 24°C, 25°C and 28°C is 22.38, 24.21 and 27.78 mmHg respectively. On that day, the relative humidity of Barisal was 65%. [RB’22] (c) What was the dew point of that day in Rangpur? 3 (d) Where will wet clothes dry faster between Rangpur and Barisal? Verify mathematically. 4 Solution (c) Dew point, θ = θ1 − G(θ1 − θ2 ) θ1 = reading obtained from dry bulb = 28° − 1.65 (28° − 26°) = 24.7°C (Ans.) θ2 = reading obtained from wet bulb (d) From ‘c’ we have obtained, dew point in Rangpur = 24.7°C Due to increase in temperature by (25 − 24)°C = 1°C, increase in saturated vapor pressure = (24.21 − 22.38) = 1.83 mm Hg ∴ Due to increase in temperature by 0.7°C, increase in saturated vapor pressure = 1.83 × 0.7 = 1.281 mm Hg ∵ Saturated vapor pressure at dew point, f = (22.38 + 1.281) = 23.661 mmHg Vapor pressure at air temperature, F = 27.78 mmHg 23.661 Relative humidity in Rangpur, R = 27.78 × 100% = 85.17% Relative humidity in Barisal, R′ = 65%. ∴ R > R′, Therefore, wet clothes will dry faster in Barisal. 08. One day the air temperature above the water of Kaptai lake was 28°C, atmospheric pressure was 105 Pa and dew point was 10.5°C. The temperature of water at the bottom of the lake was 7°C and the average density of water was 1000 kgm−3. On that day, an air bubble from the bottom of the lake reached to the surface of the water and its radius was tripled. [Saturated vapor pressure at 10°C, 11°C, and 28°C is 9.2, 9.9, and 28.5 mmHg respectively.] [Ctg.B’22] (c) Determine the relative humidity of air above the lake on that day. 3 Solution (c) Here, Due to change in temperature by (11° − 10°) = 1°C, the change in saturated vapor pressure = (9.9 − 9.2) = 0.7 mm Hg Due to change in temperature by 0.5°C, the change in saturated vapor pressure = (7 × 0.5)mm Hg = 0.35 mm Hg 17  Saturated vapor pressure at dew point or 10.5°C will be, f = (9.2 + 0.35)mm Hg = 9.55 mm Hg Saturated vapor pressure at air temperature, F = 28.5 mm Hg 𝑓 9.55 Relative humidity, R = F × 100% = 28.5 × 100% = 33.5% Relative humidity of air at the surface of the lake = 33.5% (Ans.) 09. In an experiment, the temperature of the air in Mithamain was 19°C and relative humidity is 47%. During cold wave, the temperature at that place falls down to 15°C temperature. At temperatures 7°C, 8°C & 19°C saturated water vapor pressure equals to 7.5, 8.2 and 16.5mm mercury pressure. [BB’22] (c) Determine the amount of condensed water vapor at 5°C. 3 (d) Will the dewpoint of the air contained in that place will change if the temperature of that place changes? Show with the help of mathematical analysis. 4 Solution (c) Given that, Saturated water vapor pressure at 19°C temperature, f19 = 16.5 mm Hg Relative humidity = 47% ∴ At 19°C temperature the water vapor pressure present in air, f = (16.5 × 0.47)mm Hg = 7.755 mm Hg Now, Saturated water vapor pressure at 5°C temperature, f8 −f7 8.2−7.5 f5 = f7 + (5 − 7) × = 7.5 − 2 × = 6.1 mm Hg 8−7 1 7.755−6.1 ∴ The percentage of condensed water vapor = 7.755 × 100% = 21.34% (Ans.) (d) Dewpoint is not dependent on temperature. The temperature at which the air of a place becomes saturated with water vapor is called dewpoint. Now if temperature is changed, the amount of water vapor present in the air will not change. For example: the temperature of air of a place is 15° C and the air is saturated with water vapor. Now, if the temperature is increases then the amount of water vapor present in the place will not change but the air will become unsaturated. Still the dew point will remain 15° (c) hat is why, for changing temperature the dewpoint of the air contained in that place will not change. 10. On a day in Dhaka, the temperature was 35°C and relative humidity was 50%. On the same day, the reading of a dry bulb thermometer of a hygrometer placed in Rajshahi is 25°C and the reading of a wet bulb thermometer is 19°C. [At 25°C temperature, the value of G is 1.65. At temperatures 15°C, 16°C and 25°C the pressure of saturated water vapor is 12.77 mm, 13.71 mm, 23.7 mm mercury respectively] [JB’22] (c) Determine the dewpoint of Rajshahi. 3 (d) In which place of the stem on the mentioned day will a wet cloth dry faster? Explain with mathematical logic. 4 Solution (c) Dewpoint of Rajshahi, θ = θ1 − G(θ1 − θ2 ) ∴ θ = 25° − 1.65(25° − 19°) θ1 = dry bulb temperature = 15.1°C (Ans.) θ2 = wet bulb temperature 18 (d) From ‘c’ we get, dewpoint in Rajshahi = 15.1°C For (16° − 15°) = 1°C change, change in saturated vapor pressure = (13.71 − 12.77) = 0.94 mm Hg For 0.1°C change, change in saturated vapor pressure = 0.94 × 0.1 = 0.094 mm Hg Then saturated vapor pressure at 15.1°C , f = (12.77 + 0.094) = 12.864 mm Hg And saturated vapor pressure at 25°C, F = 23.7 mm Hg 12.864 Then relative humidity in Rajshahi, R = 23.7 × 100% = 54.28% That means, Relative humidity in Rajshahi is more than that of Dhaka. That is why, clothes will dry faster in Dhaka. 11. One day in a specific area, the readings of a dry and wet bulb hygrometer were 20°C and 12.8°C respectively. At 20°C temperature glacier coefficient is 1.79. Saturated water vapor pressure at 7°C, 8°C and 20°C temperature is 7.5 × 10−3 mHgP , 8.1 × 10−3 mHgP and 17.4 × 10−3 mHgP respectively. [CB’22] (c) The dewpoint of that specific area on that day. 3 (d) Give your opinion of the weather of the specific area on that specific day mentioned in the stem with mathematical analysis. 4 Solution (c) Dewpoint, θ = θ1 − G(θ1 − θ2 ) = 20° − 1.79(20° − 12.8°) = 7.112°C ∴ So, dewpoint on that mentioned day is 7.112°C (Ans.) (d) From ‘c’ we get, Dewpoint = 7.112°C. Saturated water vapor pressure at 7°C is, f7 = 7.5 × 10−3 m Hg Saturated water vapor pressure at 8°C is, f8 = 8.1 × 103 m Hg f8 −f7 Saturated water vapor pressure at 7.112°C or dewpoint is, f7.112 = f7 + × 0.112°C 1°C 8⋅1×10−3 −7.5×10−3 = 7.5 × 10−3 + ( 1°C × 0.112) m Hg = 7.5672 × 10−3 m Hg Saturated water vapor pressure at 20°C or air temperature is, f20 = 17.4 × 10−3 m Hg f7⋅112 7⋅5672×10−3 Then, Relative humidity, R = × 100% = × 100% = 43.49% f20 17.4×10−3 Since R < 50%, So the weather in the place mentioned in the stem will be dry. 12. The temperature on a certain day is 30°C and humidity is 80%. Helen turns on the AC in the house and the temperature falls down to 21°C. The dewpoint on that day was 9.5°C [At temperatures 30°C, 21°C, 9°C & 10°C saturated water vapor pressure is 28.02 mm, 20.35 mm, 8.91mm & 9.2 mm HgP respectively.] [MB’22] (c) How much part of water vapor present in the air will condense when the temperature falls? Determine. 3 (d) Why would Helen feel comfortable when AC is turned on? Give mathematical explanation. 4 Solution (c) AT 30°C temperature, saturated water vapor pressure = 28.02 mm Humidity, f = 80% Then, at temperature 30°C pressure of water vapor present in the air is = (0.8 × 28.02)mm = 22.416 mm At 21°C saturated water vapor pressure = 20.35 mm 22.416−20.35 Then, the amount of water vapor present in air that will condense = ( 22.416 )× 100% = 9.22% (Ans.) 19 (d) In the question, it is said to determine the amount of humidity with respect to the external environment. If the temperature drops to 21°C; So, the temperature decreases with respect to outside. That is why it feels comfortable. Saturated water vapor pressure at dewpoint = Saturated water vapor pressure at 9°C temperature +(9.2 − 8.91) × 0.5 = 8.91 + 0.145 = 9.055 mm Hg Saturated water vapor pressure at dewpoint Alongside this, the relative humidity will be then, f ′ = Saturated water vapor pressure at 21° temperature × 100% 9.055 = 20.35 × 100% = 44.5% That is, the relative humidity decreases from 80% and comes within the comfortable range (40% − 60%). That is why, Helen feels comfortable by turning on the AC. 13. At a particular place, the dry bulb temperature of the hygrometer is 24°C and the dew point is 11.5°. The vapor pressure at 24°C, 12°C and 11°C is 22.38 × 10−3 m, 10.52 × 10−3 m and 9.9 × 10−3 mHg respectively. Glaisher’s factor at 24°C temperature is 1.72. [DB’21] (c) What is the reading of the wet bulb at that place? Determine. 3 (d) Mathematical analyze the weather of the above-mentioned place by determining its relative humidity and give your opinion. 4 Solution (c) We know, dew point, θ = θ1 − G(θ1 − θ2 ) Here, θ = 11.5°C; Temperature of the dry bulb, θ1 = 24°C G = Glaisher’s factor = 1.72 and reading of the wet bulb = θ2 Now, θ = θ1 − G(θ1 − θ2 ) ⇒ 11.5°C = 24°C − 1.72(24°C − θ2 ) ⇒ θ2 = 16.73256°C ∴ Reading of the wet bulb is 16.73256°C (Ans.) (d) Diffence of the vapor pressure at 1°C temperature within 11°C and 12°C = (10.52 × 10−3 − 9.9 × 10−3 )m = 6.2 × 10−4 m ∴ Saturated vapor pressure at 11.5°C = (9.9 × 10−3 + 6.2 × 10−4 × 0.5)m = 10.21 × 10−3 m Present vapor pressure, f = Dew point saturated vapor pressure ∴ f = 10.21 × 10−3 m and saturated vapor pressure at the air temperature of 24°C, F = 22.38 × 10−3 m f 10.21×10−3 m ∴ Relative humidity, R = F × 100% = 22.38×10−3 m × 100% = 45.62% ∴ The weather is roughly dry. 14. A meteorologist collected the following data for preparing the daily forecast: [RB’21] Thermometer reading from Thermometer reading from Glaisher’s factor at air Region the dry bulb the wet bulb temperature Dhaka 28.6°C 20°C 1.664 Dinajpur 32.5°C 22°C 1.625 [Saturated vapor pressure at 14°C, 16°C, 28°C, 30°C, 32°C and 34°C temperature are respectively 11.99, 13.63, 28.35, 31.83, 35.66 and 39.90 mm Hg] (c) Determine the dew point of Dhaka at that particular day. 3 (d) Based on the above information, a person would feel more comfortable in which region? Analyze. 4 20  Solution (c) Given, Glaister’s factor, G = 1.664; Reading from the dry bulb, θ1 = 28.6°C Reading from the wet bulb, θ2 = 20°C Now, dew point θ = θ1 − G(θ1 − θ2 ) = 28.6°C − 1.664(28.6°C − 20°C) = 14.2896° (Ans.) (d) Dew point in Dinajpur, θ′ = 32.5°C − 1.625(32.5°C − 22°C) = 15.4375°C saturated vapor pressure at dew point Now, relative humidity, R = saturated vapor pressure at air temperature saturated vapor pressure at 14.2896°C ∴ Relative humidity in Dhaka, R1 = saturated vapor pressure at 28.6°C (13.63−11.99) 11.99 m Hg + mm Hg×0.2896°C 12.227472 mm Hg 2°C = (31.83−28.35) × 100% = × 100% = 41.5985% 28.35 mm Hg + mm Hg×0.6°C 29.394 mm Hg 2°C (39.9−35.66) Again, Saturated vapor pressure at 32.5°C = 35.66 mm Hg + 2°C mm Hg × 0.5°C = 36.72 mm Hg (13.63−11.99) mm Hg Saturated vapor pressure at 15.4375°C = 11.99 mm Hg + 2°C × 1.4375°C = 13.16875 mm Hg 13.16875 mm Hg ∴ Relative humidity in Dinajpur, R 2 = 36.72 mm Hg × 100% = 35.8626% ∵ R 2 < R1 ∴ A person would feel more comfortable in Dinajpur. 15. [Ctg.B’21] Reading from a hygrometer Region Dry bulb Wet bulb Glaisher’s factor at air temperature Sylhet 32°C 25°C 1.63 Kuakata 33°C 27°C 1.62 Saturated vapor pressure at 20°C, 22°C, 24°C, 32°C and 34°C are respectively 17.54mm, 19.83 mm, 22.38mm, 35.66 mm and 39.90 mmHg pressure. (c) Determine the dew point of Sylhet on the day mentioned in the stem. 3 (d) Which region should be prioritized for visiting on that day? Analyze mathematically. 4 Solution (c) Here, at 32°C temperature Glaisher’s factor G = 1.63, reading from the dry bulb, θ1 = 32°C, reading from the wet bulb, θ2 = 25°𝐶. Dew point, θ = θ1 − G(θ1 − θ2 ) = 32°C − 1.63(32° − 25°C) = 20.59°C (𝐀𝐧𝐬. ) (d) Dew point in Kuakata, θ = 33°C − 1.62(33°C − 27°C) = 23.28°C saturated vapor pressure at dew point Relative humidity = saturated vapor pressure at air temperature saturated vapor pressure at 20.59°C ∴ Relative humidity in Sylhet, R1 = × 100% saturated vapor pressure at 32°C (19.83−17.54) mm Hg) 17.54mm Hg + ×0.59 18.2155 mm Hg 2 = × 100% = × 100% = 51.08% 35.66 mm Hg 35.66 mm Hg 21  39.9−35.66 Now, Saturated vapor pressure at 33°C = {35.66 + ( )× 1} mm = 37.78 mm Hg 2 (22.38−19.83) mm Hg Saturated vapor pressure at 23.28°C = 19.83 mm Hg + 2 × 1.28 = 21.462 mm Hg 21.462 mm Hg ∴ Relative humidity in Kuakata, R 2 = 37.78 mm Hg × 100% = 56.80% ∴ It can be seen that, R is low in Sylhet, hence Sylhet will be less uncomfortable. So, Sylhet should be prioritized for visiting on that day. 16. The temperature of Dhaka on a day is 35°C and the dewpoint is 19.4°C. On that time, the readings of the dry and wet bulb of a hygrometer placed in Chattogram are 35°C and 30°C respectively. [At 35°C temperature, glacier factor 1.60 & at 19°C, 20°C, 27°C and 35°C temperature, saturated vapor pressure 16.5, 17.7, 26.78 and 42.16 mm mercury respectively] [JB’21] (c) From the stem, determine the relative humidity of Dhaka on that day. 3 (d) Even if the temperature is same, where will sweat dry faster between Dhaka and Chattogram? Analyze mathematically in the light of the stem. 4 Solution (c) At 19.4°C temperature, saturated vapor pressure = 16.5 + 0.4 × (17.7 − 16.5) = 16.98 16.98 ∴ Relative humidity = × 100% = 40.28% (Ans.) 42.16 (d) From ‘c’ relative humidity in Dhaka = 40.28% Dewpoint in Chattogram θ = θ1 − G(θ1 − θ2 ) = 35 − 1.6(35 − 30) = 27°C 26.78 ∴ Relative humidity of Chattogram R = × 100% = 63.52% 42.16 ∴ In Dhaka, relative humidity is lesser, so sweat will dry up faster. 17. In a hospital, the authority tries to keep the relative humidity of the cabin of the patients under 46%. But, one day, the AC regulation unit was not working properly. The authority noticed that the readings of the dry and wet bulb hygrometer are 23°C and 15.8°C respectively; At glacier factor 23°C, the saturated vapor pressure at temperatures 1.74 ; 10°C, 11°C and 23°C are 9.2mm, 9.865mm and 21.105 mm Hg respectively. [CB’21] (c) Determine the dewpoint of that day. 3 (d) Did the hospital authority suffer from any crisis regarding the relative humidity on that day? Explain mathematically. 4 Solution (c) If dewpoint θ, θ1 − θ = G(θ1 − θ2 ) ⇒ θ = 23° − 1.74(23° − 15.8°) θ1 = 23°C ⇒ θ = 10.472°C θ2 = 15.8°C ∴ Dewpoint = 10.472°C (Ans.) G = 1.74 (d) At 10.472°C temperature, saturated vapor pressure = 9.2 + (9.865 − 9.2) × 0.472 = 9.51388 mm Hg f 9.51388 ∴ Humidity, R = F × 100% = 21.105 × 100% = 45.0788% ∴ Relative humidity is less than 46%. That is why no crisis occurred. 22  18. The temperature of a closed room is 30°C, Dewpoint is 15°C and relative humidity is 50%. The temperature outside at that time was 26°C and relative humidity was 65%. At 30°C & 26°C temperature, pressure of saturated water vapor is 31.83 mm HgP & 25.25 mm HgP respectively. At temperature 30°C , Glacier coefficient is 1.65. [Din.B’21] (c) What is the temperature of the wet bulb of the hygrometer of that room? 3 (d) If one of the windows of the room is opened, then the water vapor will flow in which direction? Explain mathematically. 4 Solution (c) We Know, θ = θ1 − G(θ1 − θ2 ) ⇒ G(θ1 − θ2 ) = (30 − 15)°C θ = 15°C; θ1 = 30°C; G = 1.65 ⇒ θ1 − θ2 = 9.091°C ⇒ θ2 = 30°C − 9.091°C Temperature of wet bulb = θ2 ⇒ θ2 = 20.909°C (Ans.) (d) Relative humidity inside house = 50% ∴ Pressure of water vapor f1 = 0.5 × 31.83 mm Hg = 15.915 mm Hg Outside house, pressure of vapor f2 = 0.65 × 25.25 mm Hg = 16.4125 mm Hg ∴ f2 > f1 ∴ Water vapor flows from the outside to the inside. 19. One day, at a place, the temperature of a dry and wet bulb in a hygrometer are 25°C & 19°C respectively and dewpoint is 14.77°C. At temperatures 15°C, 16°C and 25°C, saturated vapor pressure is 17.54 mmHg, 19.83 mmHg and 25.21 mmHg respectively. [MB’21] (c) Determine the glacier coefficient. 3 (d) If on that day the relative humidity decreases by 20%, then what type of change will occur in the dewpoint? Explain mathematically. 4 Solution (c) We know, θ1 − θ = G(θ1 − θ2 ) θ −θ 25°−14.77° θ1 = 25°C ⇒ G = θ 1−θ = 25°−19° 1 2 θ2 = 19°C ∴ G = 1.705 (Ans.) θ = 14.77°C f (d) Relative humidity, R = × 100% F 1°C temperature decrease, saturated vapor pressure decreases by (19.83 − 17.54) mmHg = 2.29 mmHg For 14.77°C temperature, F = 25.21 mmHg f = (17.54 − 2.29 × 0.23) mmHg = 17.0133 mmHg 17.0133 ∴R= 25.21 × 100% = 67.486% For 20% decrease, relative humidity = (67.486 − 20)% = 47.486% 47.486 f′ Now, 100 = 25.21 mmHg ⇒ f ′ = 11.9713 mmHg 17.54−11.9713 ∴ Dewpoint will be = 15°C − ( 2.29 )× 1°C = (15 − 2.4317)°C = 12.5683°C ∴ Dewpoint will decrease = (14.77 − 12.5683)°C = 2.2017°C 23 20. A cylinder of volume 0.6 m3 at room temperature is filled with 800 gm of methane (CH4 ) gas at a pressure of 202650 Pa at a definite location. The teacher told his students that rain is likely to happen only when the dew point of the place is 11.5°C and the relative humidity of the place is above 60%. The saturated water vapor pressure at 11°C, 12°C, 19°C and 20°C was found to be 9.84 mm(Hg), 10.52 mm(Hg), 16.46 mm(Hg) and 17.54 mm(Hg) respectively. The molecular mass of methane is 16 gmol−1. [Din.B’19] (d) Analyse mathematically if there is any possibility of rain at the location of the stem. 4 Solution (d) Temperature of that place = (292.49 − 273)°C = 19.49°C Change of vapor pressure for the change of (12 − 11)°C or 1°C temperature = (10.52 − 9.84) = 0.68mm(Hg) ∴ Change of vapor pressure for the change of 0.5°C temperature = (0.68 × 0.5) = 0.34mm(Hg) ∴ Saturated dew point at 11.5°C temperature = (9.84 + 0.34) = 10.18mm(Hg) Again, change of vapor pressure for the change of (20 − 19)°C or 1°C temperature = (17.54 − 16.46) = 1.08 mm(Hg) The change of vapor pressure for the change of 0.49°C temperature = (1.08 × 0.49 ) mm (Hg) = 0.5292mm(Hg) Saturated dew point at 19.49°C temperature = (16.46 + 0.5292) = 16.9892 mm(Hg) f 10.18 ∴ Relative humidity, R = F × 100% = 16.9892 × 100% = 59.92% ∴ There is no possibility of rain at the location of the stem. CQ Knowledge-Based Questions & Sample Answers from Previous Board Exams 01. State Charles’s Law. [Ctg.B’23] Ans: "Under constant pressure, the volume of any gas is directly proportional to its absolute temperature" - This is Charles's Law. 02. What is hygrometry? [BB’23] Ans: The branch of physics that discusses the studies of humidity and measurement is called hygrometry. 03. What is called dewpoint? [DB’23, 21, SB’23, 22, MB’22, RB’21; Ctg.B’21; BB’19; CB’19; RB’17; BB’17; RB’15] Ans: All gases that obey the fundamental postulates of the kinetic theory of gas, and also obeys Boyle’s law and Charles’s law combinedly at all temperatures and pressures are known as ideal gas. 04. What is ideal gas? [RB’23, 19; SB’ 21; MB’21; Din.B’17; Ctg.B’15; SB’15] Ans: Those gases which at all temperature and pressure does not follow Boyle’s, Charles’ and Avogadro’s law are called real gas. 05. What is saturated water vapor pressure? [SB’23, CB’23, Ctg.B’19; DB’16] Ans: The maximum pressure that the water vapor in the air of a confined place exerts at a specific temperature is called saturated water vapor pressure. 06. What is called mean free path? [JB’23, 22] Ans: The distance between two consecutive collisions of the same molecule with other molecules is called the mean free path of the molecule. 07. What is the law of equipartition of energy? [MB’23, RB’22] Ans: For any thermodynamic system in thermal equilibrium, the total energy for the system is equally distributed among the degree of freedom and the amount of energy per degree of freedom is equal. 24 08. What is real gas? [BB’22] Ans: Those gases which at all temperature and pressure does not follow Boyle’s, Charles’s and Avogadro’s law are called real gas. 09. Define relative humidity. [JB’22; DB’21; Din.B’21; Ctg.B’17; CB’17; RB’16] Ans: At a specific temperature, in a specific volume of air, the ratio of the mass of the water vapor present and the necessary mass of water vapor required to saturate that specific volume of air at that temperature is the relative humidity of that area. 10. What is absolute zero temperature? [Ctg.B’ 21; CB’21] Ans: The temperature at which the volume of a given mass of gas at constant pressure is zero is called absolute zero temperature. 11. What is a hygrometer? [SB’ 21] Ans: The instrument used for measuring the relative humidity of the air is known as hygrometer. 12. What is universal gas constant? [BB’ 21; CB’15] Ans: The amount of work done to increase the temperature of one mole ideal gas by one kelvin is called universal gas constant. 13. What is standard pressure? [BB’ 21; All B’18; Ctg.B’16; DB’15] Ans: At 45° latitude, in 273K temperature, the pressure that a dry and pure mercury column of 760mm applies is called standard pressure. 14. State Boyle’s Law. [Ctg.B’22; JB’21; JB’19] Ans: If the temperature is constant, the volume of gas of a definite mass and its pressure is inversely proportional. 15. What is absolute humidity? [Din.B’22; CB’ 21; BB’15] Ans: The mass of the air present in a unit volume of air in a specific place is called the absolute humidity of that place. 16. What is degree of freedom? [CB’22; DB’ 19; RB’17] Ans: The number of independent variables needed to express the overall condition of motion of an object is called degree of freedom. 17. What is root mean square velocity? [SB’ 17] Ans: The square root of the mean of square of velocities of all the molecules of gas is called root mean square velocity. CQ Comprehensive Questions & Sample Answers from Previous Board Exams 01. Why do we see dewdrops in winter morning? Explain. [RB’23, CB’ 21] Ans: At night, the surface of earth cools down by radiating heat and the temperature goes below dewpoint. As a result, dewdrops form. In winter morning the temperature is very low. That is why the temperature stays below the dewpoint. For this reason, we can see dewdrops in winter morning. 02. Why does fog fall in any region? Explain. [Ctg.B’23] Ans: When the temperature in the near-surface atmospheric layer decreases, the water vapor in the air condenses into tiny droplets, forming clouds. This is why dew forms. 25 03. Explain how the kinetic energy of a molecule of a gas change with temperature, through a graph. [SB’23] Ans: Kinetic energy of a gas molecule is directly proportional to absolute temperature. 3 We know, E = 2 KT 3 Here, K is a constant. So, E ∝ T 2 The graph shows that as the absolute temperature increases, the kinetic energy