Question

Fixed mass of an ideal gas contained in a 24.63 L sealed rigid vessel at 1 atm is heated from $-{73}^{0}Cto{27}^{0}C$. Calculate change in Gibb's energy it entropy of gas is a function of temperature as $S=2+{10}^{-2}T\left(J/K\right)$: (Use 1 atm L = 0.1 kJ)

• A. 1231.5 J
• B. 1281.5 J
• C. 781.5 J
• D. 0

Answer 1

Answer: (C)

781.5 J

We know that

$dG=VdP-SdT$

$=V\Delta P-{\int }_{-{73}^{o}C}^{{27}^{o}C}(2+{10}^{-2}T)dT$

$=V\Delta P-{\int }_{200k}^{300k}(2+{10}^{-2}T)dT$

At constant volume, $\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow P_2=P_1\times \frac{T_2}{T_1}=1\times \frac{300}{200}=\frac{3}{2}$

$\therefore dG=24.63\times (1.5-1.0)\times 0.1\times {10}^3-{\int }_{200}^{300}(2+{10}^{-2}T)dT$

$=781.5J$ .