Question

SO2(g) in a piston chamber kept in a constant-temperature bath at 25.0C expands from 25.0 mL to 85.0 mL very, very slowly. Assume that 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 are 0.150 mol of gas in the chamber, calculate ∆Ssys, ∆Ssurr, and ∆Suniv for the process. Note: even though the gas is not ideal in this case, you may assume that ∆U = 0 for this constant-temperature process.

Answer #1

for Vanderwaal gas (P+an^{2}/V^{2})*(V-nb)=
nRT

P= nRT/(V-nb)- an^{2}/V^{2}

From maxwell relation,
(dS/dV)_{T}=(dP/dT)_{V}

For Vanderwaal gas (dP/dT)V= nR/(V-nb)

hence (dS/dV)T= nR/(V-nb)

dS= nRdV/(V-nb)

when integrated, the entroy change= nR* ln (V_{2}-nb)/
(V1-nb)

V2= 85ml= 85/1000 L=0.085 L, V1= 25 ml= 25/1000L=0.025L

hence entropy change= 0.15*8.314* ln {( 0.085-0.15*0.05636)/ (0.025-0.15*0.05636)}=1.91 J/K

since the process is slowly carried out, the process is reversible since entropy is state function, entropy change for surr=- 1.91 J/K and total entropy change= 0

dS= (nR/(V-nb)dV

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