Question

7) A 1.20 mole sample of an ideal gas at 28.0°C expands isothermally from an initial volume of 24 dm to a final volume of 75 dm3. Calculate w for the following processes: (a) an expansion against an external pressure of 0.5x 105 Pa (b) a reversible expansion. Please give these quantities in joules. This may involve unit conversion(s).

Answer 1

Given n = 1.2 mol, T = 28 degree C = 301.15 k V_initial = 24 dm^3 = 0.024 m^3 V_final = 75 dm^3 = 0.075 m^3 vert 1 dm = 0.1m rightwardsdoublearrow 1 dm^3 = 0.001 m^3 (a) The work done at constant pressure is given by w = p Delta v Delta v rightarrow charge in volume Delta V = V_final - V_initial = 0.051 m^3 W = 0.5 times 10^5 pa (0.051 m^3) W = 2550 J (b) In reversible process, pressure is not constant but constantly changing. The gas passes through equilibrium states in which pressure is defined and it is given by the ideal gas equation pV = nRT rightwardsdoublearrow p = nRT/v The work done is W = integral__V_initial^V_final Pdv = integral_V_initial^V_final nRT/v dv \r\n In isothermal reversible process, T is constant and can be taken out of the integration along with n and R W_isothermal^rev = nRT integral_V_initial^V_final dv/v = nRT ln(V_final/V_initial) W_isothermal^rev = 1.2 mol times 8.314 (J k^-1 mol^-1) times 301.15 k times ln (0.075 m^3/0.024 m^3)