Question

A sample of an ideal gas is expanded to twice its original volume of $1m^3$ in a reversible process for which $P=\alpha v^2$ where α=5 $atm/m^6$ . If $C_{v,m}=20Jmo{l}^{-1}{K}^{-1}$, determine the approximate molar change in entropy for the process.

• A. $4Jmo{l}^{-1}{K}^{-1}$
• B. $47.35Jmo{l}^{-1}{K}^{-1}$
• C. $40Jmo{l}^{-1}{K}^{-1}$
• D. $4Jmo{l}^{-1}{K}^{-1}$

Answer 1

The correct option is B $47.35Jmo{l}^{-1}{K}^{-1}$

We know formula for molar entropy change is :
$\Delta S=n{C}_{v}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$
$V_{1}and{V}_2$ are given and we have to calculate temperature in terms of pressure by using ideal gas equation
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
or $\frac{T_2}{T_1}=\left(\frac{P_2}{P_1}\right)\left(\frac{V_2}{V_1}\right)$
P = $\alpha V^2$
So $P_1=\alpha V_1^2$ and $P_2=\alpha V_2^2$ put up in eq. (1), we get
$\frac{T_2}{T_1}=\left(\frac{\alpha V_2^2}{\alpha V_1^2}\right)\left(\frac{V_2}{V_1}\right)$
$\frac{T_2}{T_1}=(\frac{V_2^3}{V_1^3}$)
where $\frac{V_2}{V_1}=\frac{2}{1}$
$\frac{T_2}{T_1}=8$
$\Delta S=n{C}_{v}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$ ( n=1)
$\Delta S=1\times 20ln8+1\times Rln2$
$\Delta S=47.35Jmo{l}^{-1}{K}^{-}1$