Unit VIII: Behavior of Gases PDF

# Unit VIII: Behavior of Gases Out of the three states of matter the simplest one is the gaseous state. Gases show maximum regularity in their behavior irrespective of their nature. Some extremely useful generalizations have been deduced from the behavior of gases which are known as gas laws. ## 1...

# Unit VIII: Behavior of Gases Out of the three states of matter the simplest one is the gaseous state. Gases show maximum regularity in their behavior irrespective of their nature. Some extremely useful generalizations have been deduced from the behavior of gases which are known as gas laws. ## 1. Mass The gases do possess mass. The mass of the gas can be experimentally determined by weighing the container in which it is enclosed followed by taking the weight of empty container. The difference in the two weights gives the mass of the gas. The mass of the gas is related to the number of moles as: $n = \frac{m}{M}$ Where, * n = number of moles * m = mass of gas in grams * M = molecular mass of the gas. ## 2. Volume The volume of the gas is the space occupied by its molecules under particular set of conditions. Since gases occupy the entire space available to them, therefore, the measurement of the gas volume simply requires the measurement of the volume of the container in which the gas is enclosed. **Units of volume:** Generally, volume is expressed in terms of litres (L) or cubic metres (m³) or cm³ or dm³. The relationships between different units are: 1 m³ = 10³L = 10³dm³ = 10⁶cm³ ## 3. Pressure A gas enclosed in a vessel exerts outward force on its walls. The outward force per unit area of the walls is termed as gas pressure. ### Atmospheric pressure and its measurement It is a well known fact that the earth is surrounded by a thick blanket of air called atmosphere. The pressure exerted by the gases of the atmosphere on the surface of the earth is called atmospheric pressure. The atmospheric pressure can be measured by a simple device called barometer. A simple barometer can be made by filling mercury in a tube (longer than 76 cm) closed at one end and inverting it in an open vessel containing mercury. The mercury level in the tube adjusts itself and stands approximately 76 cm above the level of mercury in the open vessel. The height of mercury column in the tube gives a measure of atmospheric pressure. The height of the column decreases when the pressure of the atmosphere decreases whereas the height of the column increases with the increase in the atmospheric pressure. **Unit of pressure:** The SI unit of pressure is newton per square meter (Nm²) which is also called Pascal (Pa). It may be defined as the pressure exerted when a force of 1 newton (1 N) acts on 1 m area. The numerical value of 1 atmosphere pressure = 101.325kPa. **Measurement of gas pressure:** The instrument used for the measurement of gas pressure is called manometer. It consists of U tube partially filled with mercury. One limb of the tube is shorter than the other. The shorter limb is connected to the vessel containing the gas whereas the longer limb is open as shown in Fig. 3.3. The mercury in the long tube is subjected to the atmospheric pressure while mercury in the shorter tube is subjected to the pressure of the gas. There are three possibilities as described below: (i) If the level of Hg in the two limbs is same, then gas pressure = atmospheric pressure (Pa) (ii) If the level of Hg in the longer limb is higher, then gas pressure = Pa + (difference between the two levels) = Pa + h (iii) If the level of Hg in the shorter limb is higher, then gas pressure = Pa-(difference between the two levels) = Pa-h ## 4. Temperature In general, temperature may be defined as degree of hotness. One of the common temperature-measuring device is thermometer. As the zero in Celsius scale is arbitrarily fixed, therefore, it is possible to have temperature below the freezing point. It may appear that Celsius scale can be extended to negative temperatures indefinitely, but experimental behavior of gases shows that temperatures below -273.15°C are impossible to be attained. # The Gas Laws Certain generalisations were developed from the quantitative studies of the behavior of gases. These generalisations are known as gas laws. Let us study some of the gas laws in details. ## Boyle's Law This law describes the pressure-volume relationship of gases at constant temperature. It was given by Robert Boyle (1662) and is known after his name as Boyle's law. The law states that: The volume of a given mass of a gas is inversely proportional to its pressure at constant temperature. Mathematically, the law may be expressed as: (Temperature and mass constant) $V \propto \frac{1}{P}$ $V = K \frac{1}{P}$ $PV = K$ = constant. Thus, another statement of Boyle's law may be given as follows: For a given amount of the gas, the product of pressure and volume is constant at constant temperature. Let V₁ be the volume of a given mass of the gas having pressure P₁ at temperature T. Now if the pressure is changed to P2 at the same temperature, let the volume changes to V2. The quantitative relationship between the four variables P1, V1, P2 and V2 is: P₁V₁=P₂V₂ (Temperature and mass constant) The experimental verification of the law can be carried out by measuring the values of volumes of a given mass of a gas at different pressures keeping the temperature constant. In each case the product PV is found to be constant. The values of pressures and volumes of air at constant temperature are given in table below. | Pressure P (atm.) | Volume V (litres) | PV (litre atm.) | |---|---|---| | 0.20 | 112.0 | 22.4 | | 0.25 | 89.2 | 22.3 | | 0.35 | 64.2 | 22.47 | | 0.40 | 56.25 | 22.50 | | 0.60 | 37.40 | 22.44 | | 0.80 | 28.1 | 22.48 | | 1.00 | 22.4 | 22.40 | ### Graphical representation of Boyle's Law The law can also be illustrated by means of pressure-volume curves. From the plot of V vs P at a particular temperature, we can infer that P increases when V decreases. The plot of PV vs P, at particular temperature indicates that the PV values remain constant in spite of regular increase in P. The curve obtained by plotting PV against P at a particular temperature is called isotherm. The higher curve corresponds to higher temperature. The variation of P against $\frac{1}{V}$ gives a straight line indicating that $\frac{1}{V}$ regularly increases with the increase in P. ## Practical Importance of Boyle's Law The Boyle's law expresses in a quantitative manner the important experimental fact that gases are compressible. When a given mass of gas is compressed, the same number of molecules occupies a smaller space. This means that the gas becomes denser. **For example:** a) Air at the sea level is dense because it is compressed by the mass of air above it. However, the density and pressure decreases with increase in altitude. The atmospheric pressure at Mount Everest is only about 0.5 atm. (The decrease in pressure at night) altitudes causes' altitude sickness (sluggish feeling, headache, etc.) due to decrease in the oxygen intake in each breath. b) The inside of jet airplanes, which normally fly at about 10,000 m are specially maintained at normal pressure and they are also equipped with emergency oxygen supply in case the pressure falls. **Solution.** Here, P₁ = 1 atm. V₁ = 3.15 L P2 = 3.5 atm. V2 =? From Boyle's law equation : P₁V₁ = P₂V₂ $V2 = \frac{P1V1}{P2} = \frac{1.00 atm. x 3.15 L}{3.5 atm.} = 0.90 L.$ **Example 1.** What is the volume of a sample of oxygen at a pressure of 3.5 atm. if its volume at 1 atm. is 3.15 L at the same temperature? **Example 2.** A gas occupies a volume of 2.5 L at 9 x 10⁵ Nm². Calculate the additional pressure required to decrease the volume of the gas to 1.5 L, keeping temperature constant. **Solution.** Here, P₁ = 9 x 10⁵ Nm² P2 =? V1 = 2.5 L V2 = 1.5 L From Boyle's law equation: P₁V₁ = P₂V₂ $P2 = \frac{P1V1}{V2} = \frac{9.0 x 10⁵ Nm⁻² x 2.5 L}{1.5 L} = 15 x 10⁵ Nm⁻²$ The additional pressure required = (15 x 10⁵ - 9 x 10⁵) Nm⁻² = 6 x 10⁵ Nm⁻². **Review exercises** 1. A sample hydrogen gas occupies 10 L at 190 mm pressure. What would be the volume of this sample at 1 atm pressure, if the temperature is kept constant? Ans. 2.5 L. 2. A sample of gas occupies 100 L at 1 atm pressure and at 0°C. If the volume of the gas is to be reduced to 5 L at the same temperature, what additional pressure must be applied? Ans. 19 atm. ## Charles's Law This law describes the relationship between volume and temperature of gases at constant pressure. It was put forward by the French chemist Jacques Charles in 1787 and was further developed in 1802 by Joseph Gay Lussac. This law can be slated as: The volume of a given mass of a gas increases or decreases by 1/273 of its volume at 0°C for each degree rise or fall of temperature, provided pressure is kept constant. If V₀ is the volume of given mass of the gas at 0°C Then, V₁, the Volume of the gas at 1°C= V₀ + $\frac{V₀}{273}$ = V₀ (1+$\frac{1}{273}$) V2, the volume of the gas at 2°C = V₀ (1+$\frac{2}{273}$) V₁, the volume of the gas at t°C = V₀ (1+$\frac{t}{273}$) ### Absolute scale of temperature By carrying out the similar calculations it can be shown that the volume of the gas below 0°C will be less than V₀. For example, the volume of the gas at -t°C `V-₁ = Vo (1-t/273)` Thus, decrease of temperature results in the decrease in the volume of the gas and ultimately, the volume should become zero at -273°C. It means that any further lowering of temperature is impossible because it would correspond to negative volume which is meaningless. Hence, an important conclusion can be drawn from the above discussion that the lowest possible temperature is -273°C. This lowest possible temperature at which all the gases are supposed to occupy zero volume is called Absolute zero. A scale of temperature based upon this choice of zero is called absolute scale of temperature. Since this scale was suggested by the British scientist Lord Kelvin, it is also known after his name as Kelvin scale of temperature. Careful measurements have revealed that absolute zero of temperature is 273.15°C. Temperatures on the Kelvin scale are indicated by writing the letter K. By convention, the degree sign (°) is not used while expressing temperatures on Kelvin scale. For example, -273.15°C=0K The relationship between Kelvin scale and Celsius scale is: T= t+273.15 Where, 'T' is the temperature on Kelvin scale while't' is the temperature on Celsius scale. While solving numerical problems, the temperature in Celsius scale is converted into Kelvin scale by addition of 273 instead of 273.15 for the sake of simplicity. It is worthwhile to mention here, that our conclusion that gases occupy zero volume at 0 K cannot be realized in actual practice because all the gases condense to liquids and solids before this temperature is reached. However, Kelvin scale is quite significant for scientific work and can be justified by thermodynamic arguments. For this reason it is also called sometimes, thermodynamic scale of temperature. ### Alternative statement of Charles's law We have already derived the relationship between the volume of a given mass of the gas at t°C (V₁) and that at 0°C (V₀). (273 + V₁ = Vo (1+$\frac{t}{273}$) = Vo ( $\frac{273 + t}{273}$ ) `= Vo $\frac{T₁}{T₀}$ Where T is corresponding temperature on Kelvin scale $\frac{V₁}{V₀}$ = $\frac{T₁}{T₀}$ V₁ :. T₁= constant (at constant P) V T₁= constant (at constant P) Thus, Charles's law may be stated in an alternate way as: The volume of the given mass of a gas at constant pressure is directly proportional to the temperature on Kelvin scale. Let V₁ be the volume of a certain mass of a gas at temperature T₁ and at pressure P. If temperature is changed to T2 keeping pressure constant, the volume changes to V2. The relationship between four variables V1 T1, V2 and T2 is: $\frac{V₁}{T₁}$ = $\frac{V₂}{T₂}$ (Pressure and mass constant) The experimental verification of the law can be done by measuring the volumes of the given mass of a gas at different temperatures keeping the pressure constant. In each case the ratio V/T comes out to be constant. The volume-temperature data for a certain mass of a gas is given in table below | Temperature °C (t) | Volume | (V/T) | |---|---|---| | - 50 | 223 | 1.00 | | 0 | 273 | 1.00 | | 50 | 323 | 1.00 | | 100 | 373 | 1.00 | | 150 | 423 | 1.00 | ### Graphical representation of Charle's Law The law can also be illustrated by volume-temperature curves. The above graph gives a plot of Volume of a given mass of a gas against temperature at constant pressure. It also depicts that volume of a gas at constant pressure is a linear function of its temperature. If we plot a graph of variation of volume with temperature (expressed on kelvin scale) at constant pressure, we get the graph as below. The curve obtained by plotting volume (V) vs temperature at particular pressure is called isobar. ## Practical Importance of the Charle's law a) The use of hot-air balloons in sports and for meteorological observations is an interesting application of the Charle's law. According to Charle's law, the gases expand on heating. Since mass of the gas is unchanged, therefore, larger volume corresponds to lower density. Thus, hot air is less dense than cold air. This causes the hot air balloon to rise by displacing the cooler air of the atmosphere. b) The gases are liquefied at low temperature based on the principle of Charles's law. ## Gay Lussac's Law The pressure of a given mass of a gas increases or decreases by 1/273 of its pressure at 0°C for each degree rise or fall of temperature, provided volume is kept constant. $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ ## Temperature vs Pressure (Volume is constant - Isochore) ## Avogadro's Law This law relates the volume of a gas to the number of molecules at constant temperature and pressure. It was given by Amedeo Avogadro in 1811. It states that equal volumes of all the gases under similar conditions of temperature and pressure contain equal number of molecules. For example, 22.4 litres of all the gases at 273 K and 1 atm. pressure contain 6.02 x 10²³ molecules. It follows, therefore, that volume of the gas is directly proportional to the number of molecules. `V ∝ N` (Temperature and pressure constant) The number of molecules (N) of any gas is directly proportional to its number of moles (n). Thus, `V x n` (Temperature and pressure constant) ## Ideal gas equation We have studied different gas laws in the previous sections. The combination of these laws namely; Boyle's law, Charle's law and Avogadro's law leads to the development of the equation which relates four variables pressure, volume, absolute temperature and number of moles. The equation so formulated is called gas equation. In order to develop gas equation, let us consider a certain quantity of gas occupying a volume V₁ at temperature T₁ and pressure P1. Let the pressure and temperature be changed to P, and T2 respectively, so that the volume changes to V2. Suppose this change from initial state to final state has been brought about in two stages as shown below: The above equation is quite useful for calculating one of the variables if the values of other variables are known. Another important conclusion that can be drawn from equation 3 is that the ratio of PV to the absolute temperature T for a given quantity of a gas is constant. That is, $\frac{PV}{T}$ = constant = KT The value of the constant K depends upon the amount of the gas. In order to make the equation independent of the amount of the gas we take the help of Avogadro's law according to which, the volume of the gas at constant temperature and pressure is proportional to its number of moles (n). This means that K is directly proportional to the number of moles (n). Thus, K ∝ n Or K = nR where R is constant of proportionality which is independent of the amount as well as nature of the gas and is known as universal gas constant. It leads to the conclusion that $\frac{PV}{T}$ = R PV = nRT ### Alternative derivation of gas law: According to Boyle's law: V ∝ $\frac{1}{P}$ (at constant T and n) .....(i) According to Charle's law: V ∝ T (at constant P and n) According to Avogadro's law: V ∝ n (at constant T and P) .....(iii) Combining (i), (ii) and (iii) PV ∝ nT PV = nRT where, R = constant of proportionality and is known as universal gas constant. ### Numerical value of R The numerical value of R depends upon the units in which pressure and volume are expressed. Let us calculate the value of R when pressure and volume are expressed in different units. (i) In SI units, pressure is expressed in Nm² and volume in m³. Then numerical value of R in SI units can be calculated as follows: From general gas equation, R = $\frac{PV}{nT}$ Substituting the values for one mole of gas at S.T.P., R = $\frac{(101325 Nm² )(22.4 x 10⁻³ m³)}{(1 mol)(273.15 K)}$ = 8.31 JK⁻¹ mol⁻¹. (ii) When pressure is expressed in atmosphere and volume in litres the value of R comes out to be in litre-atmosphere per degree per mole. **Standard Temperature and Pressure** Substituting these values for 1 mole of gas at S.T.P., R= $\frac{1(atm) x 22.4 (litres)}{(1 mol) x 273.15(K)}$ = 0.0821 litre atm K⁻¹ mol⁻¹. Since volume of a given mass of a gas depends on temperature and pressure both, it is necessary to specify the values of P and T when the value of V is stated. In general, the comparison of the volumes of different gases is made with reference to standard temperature and pressure. For gases, the standard temperature is taken as 0°C (273.15 K) and standard pressure as 1 atm (101.325 kPa) or 760 mm of Hg. The conditions are quite often abbreviated as S.T.P. meaning standard temperature and pressure or N.T.P. meaning normal temperature and pressure. In brief, S.T.P. or N.T.P. refer to: Temperature = 0°C or 273.15 K Pressure = 1 atm. or 760 mm of Hg or 760 torr or 101.325 kPa. Let us now proceed to solve some numerical problems. ## Problems Based Upon Equation : $\frac{P₁V₁}{T₁}$ = $\frac{P₂V₂}{T₂}$ **Example** A sample of nitrogen gas occupies a volume of 320 cm³ at S.T.P. Calculate its volume at 66°C and 0.825 atm pressure. **Solution.** Here, P₁ = 1.00 atm P2 = 0.825 atm V₁ = 320 cm³ V2=? T₁ = 273 K T2=66°C = 66+273 = 339 K According to the gas equation, $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ $V2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{1 x 320 x 339}{273 x 0.825} = 482 cm³.$ ## Problems Based Upon Ideal Gas Equation: (PV = nRT) **Example 1.** How many moles of oxygen are present in 400 cm³ sample of the gas at a pressure of 760 mm of Hg at a temperature of 27°C. (The value of R is given to be 8.31 kPa dm³ K⁻¹ mol⁻¹) **Solution.** Since value of R is given in SI units. Therefore, it is necessary to express the pressure and volume in kilo pascals and dm³ respectively. Thus, P = 760 mm = 101.3 kPa V = 400cm³ = 400 x 10⁻³dm³ = 0.4 dm³ T = 27°C = 27+273 = 300K n =? According to ideal gas equation, PV = nRT $n = \frac{PV}{RT} = \frac{101.3 X 0.40}{8.31 x 300} = 0.0162 = 1.62 x 10⁻² mol.$ **Example 2.** A discharge tube containing nitrogen gas at 25°C is evacuated till the pressure is 2 x 10⁻² mm. If the volume of discharge tube is 2 litres. Calculate the number of nitrogen molecules still present in the tube. (R = 0.0821 L-atm mol⁻¹ K⁻¹). **Solution. Step I. Calculation of number of moles of nitrogen** Here, P = 2 x 10⁻² mm = 2 x 10⁻² x $\frac{1}{760}$ atm V= 2 L T=25°C = 298 K n =? According to ideal gas equation, PV = nRT $n = \frac{PV}{RT} = 2 x 10⁻² x \frac{1}{760} x \frac{0.0821 x 298}{2} = 2.15 x 10⁻⁶ mol$ **Step II. Calculation of number of molecules** 1 mole of nitrogen contains = 6.02 x 10²³ molecules 2.15 x 10⁻⁶ mol of nitrogen contain = 6.02 x 10²³ x 2.15 x 10⁻⁶ molecule. = 1.29 x 10¹⁸ molecules. **Example 3.** Calculate the pressure of 1 x 10²² molecules of sulphur dioxide when enclosed in a vessel of 2.5 L capacity at the temperature of 27°C. **Solution. Step I. Calculation of number of moles** The number of moles 'of SO₂ = $\frac{1 x 10^{22}}{6.02 x 10^{23}}$ = 0.166 x 10⁻¹ mol = 1.66 x 10⁻² mol. **Step II. Calculation of pressure** Here, V= 2.5 L T=27°C = 300 K n = 1.66 x 10⁻² mol. Applying ideal gas equation, PV = RT $P = \frac{nRT}{V} = 1.66 × 10⁻² x 0.0821 x 300$ = 1.635 x 10¹ atm. **Review exercises** (i)The volume of a gas is 10.00 L at N.T.P. Find the volume of the gas at 100°C and 5.0 atm pressure. Ans. 2.73 L. (ii) What is the pressure exerted by 4.0 moles of nitrogen in a 20.0 L vessel at 27°C. Ans. 4.9 atm. # DALTON'S LAW OF PARTIAL PRESSURES This law describes the relation between the pressure of the mixture of non-reacting gases enclosed in a vessel to their pressure exerted by a mixture of two or more non-reacting gases enclosed in a definite volume, is equal to the sum of the individual pressures which each gas would exert if present alone in the same volume. The individual pressures of gases are known as partial pressures. If P is the total pressure of the mixture of non-reacting gases at temperature T and volume V, and P1, P2, P3... represent the partial pressures of the gases, then ` P = P₁ + P₂ +P₃ + ....` (T, V are constant) The law can be illustrated by considering the following example. Suppose we have three containers of capacity 1 litre each; one containing x moles of nitrogen, the other y moles of oxygen and the third having a mixture of x moles of nitrogen and y moles oxygen. All the three containers, are kept at the same temperature. Now, if the manometer, attached to first container shows a pressure P₁ and that attached to second container shows a pressure P2. Then the pressure in the third container is P₁ + P2. **Utility of Dalton's law.** This law is useful in calculating the pressure of the gas collected by the displacement of water. The gas being collected over water also contains water vapours. The observed pressure of the moist gas is equal to the sum of the pressure of the dry gas and the pressure of the water vapours. The pressure of the water vapours is constant at a particular temperature and is known as aqueous tension at that temperature. Thus, `Pobserved = Pgas + aqueous tension` `Pgas = Pobserved - aqueous tension` Let us now apply, Dalton's law to solve some numerical problems. **Example 1.** A certain quantity of a gas occupies a volume of 0.1 L when collected over water at 15°C and a pressure 0.95 atm. The same gas occupied a volume of 0.085 L at S.T.P. in dry conditions. Calculate the aqueous tension at 15°C. **Solution.** Let the aqueous tension at 15°C be p atm. Thus, pressure of the dry gas at 15°C = (0.95 -p) atm. Now, P1 =(0.95-p) V₁ = 0.1 L T₁= 15°C=288K P2 = 1 atm V2=0.085 L T2 = 273 K According to gas equation, `$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$` `P_1 = $\frac{P_2V_2T_1}{V_1T_2}$` Substituting the values, `0.95-p = $\frac{1 x 0.085 × 288}{0.1 x 273}$ = 0.89` `p = 0.06 atm.` ## Kinetic theory of Gases This theory explains observed behavior of gases in comparison to gas laws based on experimental observations. Main postulates are: 1. All gases consist of very large number of tiny particles called molecules that are in constant motion. 2. Gas molecules are perfectly round, very hard and separated by large distances. Actual volume is negligible compared to total volume. 3. Gas molecules collide with one another and walls of the container. Pressure exerted by a gas is due to collisions of molecules to walls of container. 4. Collisions between molecules are perfectly elastic i.e. there is no loss of energy due to collisions. 5. Distance between gas molecules being large, there is no effective force of attraction or repulsion present between them. 6. Gas molecules move freely in all directions and are said to have a random motion. 7. Average kinetic energy of gas molecules is proportional to absolute temperature of gas. 8. There is no effect of gravity on gas molecules.