Question

Fixed amount of an ideal gas contained in a sealed rigid vessel $(V=24.6litre)$ at $1.0$ bar is heated reversibly from ${27}^{o}C$ to ${127}^{o}C$.

Determine change in Gibb's energy (in Joule) if entropy of gas $S=10+{10}^{2}T\left(J/K\right)$.

• A. $-530$ J
• B. $+530$ J
• C. $530$ kJ
• D. None of the above

Answer 1

The correct option is A $-530$ J

The relationship between the Gibbs free energy, enthalpy and entropy is
$G=H-TS=U+PV-TS$
$dG=dU+PdV+VdP-TdS-SdT$
$w=0,dV=0,dV=dq=TdS$
Hence,
$dG=TdS+VdP-TdS-SdT$
$dG=VdP-SdT\Rightarrow \Delta G=V\Delta P-\int SdT$......(i)
$VdP=V(P_2-P_1)$......(ii)
$\frac{P_2}{T_2}=\frac{P_1}{T_1}\Rightarrow \frac{P_2}{400}=\frac{1}{300}\Rightarrow P_2=\frac{4}{3}\times 1bar$
Substitute value of $P_2$ in equation (ii)
$VdP=24.6(4/3-1)=8.2L-atm=820J$......(iii)
$\int SdT=2{\int }_{T_1}^{T_2}(10+0.01T)dT=10(T_2-T_1)+0.005(T_2^2-T_1^2)$
$SdT=10\left(100\right)+0.005({400}^2-{300}^2)=1350$.....(iv)
Substitute (iii) and (iv) in (i)
$\Delta G=820-1350=-530J$
Hence, the change in Gibbs energy is $-530$ J.