Question

A sample of a fluorocarbon was allowed to expand reversibly and adiabatically to twice its volume. In the expansion the temperature dropped from 298.15 to 248.44 K. Assume the gas behaves perfectly. Estimate the value of $C_{V_{1}m}$.

• A. $C_{V_{1}m}=31.6J{k}^{1}mo{l}^1$
• B. $C_{V_{1}m}=3.16J{k}^{1}mo{l}^1$
• C. $C_{V_{1}m}=316J{k}^{1}mo{l}^1$
• D. None of these

Answer 1

The correct option is A $C_{V_{1}m}=31.6J{k}^{1}mo{l}^1$

Process reversibly adiabatic
$T_1=298.15K{V}_2=2V_1$

$T_2=248.44KP{v}^{\gamma }=KPV=nRT$

$\frac{T}{V}\cdot V^{\gamma }K{T}_1{V}_1^{\gamma 1}=T_2{V}_2^{\gamma 1}$

$\left(\frac{T_1}{T_2}\right)={\left(\frac{V_2}{V_1}\right)}^{\gamma -1}\left(\frac{298.15}{248.44}\right)=2^{\gamma -1}$

$1.2=2^{\gamma -1};log1.2=log2.(\gamma -1)$

$\gamma -1=\frac{log1.2}{10g2}\gamma -1=0.263$

Now $n{C}_v(T_2-T_1)=\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}$

$C_{V_{1}m}=\left(\frac{R}{\gamma -1}\right)=\frac{nR(T_2-T_1)}{(\gamma -1)}$

$C_{V_{1}m}=\frac{8.314}{0.263}{C}_{V_{1}m}=31.61J{k}^{1}mo{l}^1$