States of Matter PDF

States of Matter Behaviour of Gases  States of Matter Matter can exist in three different states – gas, liquid and solid Condition of any of these states of matter can be completely specified by specifying its: ❖Pressure (P), ❖Temperature (T), ❖a...

States of Matter Behaviour of Gases  States of Matter Matter can exist in three different states – gas, liquid and solid Condition of any of these states of matter can be completely specified by specifying its: ❖Pressure (P), ❖Temperature (T), ❖and Molar volume (Vm). Volume V Molar volume = ) i.e. Vm = no. of moles n  Gases ❑A fluid ❑occupies the whole volume of its container. ❑Molecules of a gas are widely separated and are in continuous random motion.  Liquids A fluid that takes the shape of its container Does not necessarily occupy the whole volume. Liquid molecules are closer together than in a gas. Molecules of a liquid also are in motion but in a restricted manner.  SOLIDS Solid molecules are tightly packed A solid has a defined & rigid shape, irrespective of its container shape.  An Equation of State Condition of any of these states of matter can be completely specified by specifying its: ❖Pressure (P), ❖Temperature (T), ❖and Molar volume (Vm). Liquids and solids Commonly use a single value of density or molar volume to represent their behaviour over a wide range of T & P Results in no significant loss of accuracy Gases Density and molar volume of gases vary significantly with T & P. So cannot use a single value of these properties over a wide range of T & P. To solve problems that involve gases, need to use an equation which relates Vm to T & P. Such an equation is termed “equation of state” (EOS).  Ideal Gas Laws The pressure, (p), volume (V), temperature, (T) and number of moles, (n) of an ideal gas are related by the equation: pV=nRT R = universal gas constant = 8314 J K-1 mol-1 (OR: 8.314 kJ K-1 kmol-1) ❖This is the simplest equation of state ❖ A gas that obeys this equation exactly at all T & P is called “an ideal gas” or a “perfect gas”. ❖A real gas is one that does not obey this equation except in the region of low P ❖As pressure decreases, all gases obey the ideal gas law more closely  Worked Example 1 Calculate the pressure exerted by 0.95 g of ammonia gas in a container of volume 300 mL and at a temperature of 300C. Mol mass of ammonia (NH3) = 17.03 g mol-1 R= 8.314 J mol-1 K-1. Express your answer in S. I. unit.  Solution to Worked Example 1 Given data: Mass = 0.95 g, volume = 300 mL, T = 300C, M = 17.03 g mol-1 To find: Pressure, P (Nm-2) OR (Pa) Step 1: Express given data in S.I. unit Vol, V = 300 mL = 300 x 10-06 m3 = 3 x 10-4 m3 T = 30 0C = (30+273.15)K = 303.15 K Step 2: From Ideal gas law: pV = nRT   P = nRT , but n is unknown V Step 3: Calculate n from given information n = mass 0.95  g   = 0.056 mol = Mol mass 17.03  g mol−1  Step 4: now calculate P:  P = nRT =  0.95  x 8.314x303.15 = 468657.2 Nm − 2 (Pa ) V  17.03  −4 3x10  Boyle’s Law (Effect of varying pressure at constant T & n) From the ideal gas equation: pV= nRT When n and T are constant the ideal gas equation becomes: P= (nRT ) = cons tan t V V OR : P  1 V At constant temperature, the volume of a fixed amount of gas is inversely proportional to the pressure.  Worked Example 2 Calculate the final pressure of a sample of Nitrogen initially at 1500 kPa and 100 mL, which was allowed to expand to 104 cm3 (OR: mL) at a constant temperature of 250C.  Explanation of Boyle’s law ❖ Reduce the V1 by half to V2 , then no of molecules in V2 is doubled. ❖ So, more molecules strike the walls of the container in a given time, ❖ so the average force they exert is doubled; ❖ hence the pressure is doubled, BUT pV remains constant. Isotherms of an ideal gas ❖A graph of P against V, for different T ❖At any point on the isotherm, T is constant and P varies inversely as V. Gay-Lussac’s law (effect of varying T at constant P & n) Worked Example 3: To what temperature must a 10-3 m3 sample of an ideal gas be cooled from 250C in order to reduce its volume to 10-4 m3? Assume pressure remains constant.  Avogadro’s law From the ideal gas law: pV = nRT When P & T are constant then:  RT  V = n   P  i.e. V n This equation implies that the volume of an ideal gas is proportional to the no. of moles of gas present. Avogadro’s law: Equal volumes of gases at the same T and P contain equal number of molecules. ( ) So we define: molar volume of gas, Vm, as: Vm = V n = RT = constant P i.e. the molar volume is the same for all gases at the same constant T and P  Mixture of ideal gases Assumption: No reaction between gases in the mixture. Dalton’s law: the pressure exerted by a mixture of ideal gases is the sum of the pressures exerted by each gas occupying the same volume alone. Consider a mixture of Nitrogen and Oxygen, from ideal gas law:  RT  PO 2 = nO2    V   RT  PN 2 = n N2    V   PT = PO 2 + PN 2 = RT V ( nO2 + n N2 ) For a mixture containing gases: A, B, C,....in a container, the total pressure is: NT PT = PA + PB + PC +......... =  Pi i=A  Mole fraction Mole fraction, xA: of molecules of a gas “A” in a mixture, is the no of molecules of the gas “A” expressed as a fraction of the total number of molecules present in the volume.