Question

SO2 in a piston chamber kept in a constant-temperature bath at 25.0°C expands from 25.0 mL to 75.0 mL very, very slowly. A ssume SO2 behaves as a van der Waals gas, and its van der Waals parameters are a = 6.714 atm* L2/mol2 and b = 0.05636 L/mol. If there is 0.00100 mole of ideal gas in the chamber, calculate Delta Ssys, Delta Ssurr, and Delta Suniv for the process.

Answer 1

For Vanderwall gas P =nRT/(V-nb)- an²/V²                    

(dS/dT)_V= CV/T, (dS/dV)_T= (dP/dT)_V = nR/(V-nb)

S= f(T,V), dS= (dS/dT)_V dT +(dS/dV)_T dV = (CV/T)dT + nR/(V-nb)dV

For isothermal process,

deltaS= nR* ln (V2-nb)/ (V1-nb)

=0.001*8.314*ln (0.075-0.001*0.05636)/(0.025-0.001*0.05636)=0.0091 J/K

This is the entropy change of system.

Since the process is reversible, entropy change ( total)=0

Entropy change of system+ entropy change of surroundings= 0

Entropy change of surroundings = -0.0091 J/K