Question

A sample of fluorocarbon was allowed to expand reversibly and adiabatically to twice its volume. In the expansion, the temperature dropped from 298.15 K to 248.44 K. Assume the gas behaves perfectly. Estimate the value of Cv.

Answer 1

Let $V$ be the initial volume.

Therefore,

$V_f=2V$

$T_i=298.15K$

$T_f=248.44K$

In an adiabatic process-

$t{V}^{\gamma -1}=\text{constant}$

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

$\frac{298.1}{248.44}={\left(\frac{2V}{V}\right)}^{\gamma -1}$

$1.2=2^{\gamma -1}$

$\Rightarrow (\gamma -1)\log 2=\log 1.2$

$\Rightarrow gamma-1=\frac{0.0791}{0.3010}$

$\Rightarrow \gamma =1+0.262=1.262$

As we know that,

$C_v=\frac{R}{\gamma -1}$

$\therefore C_v=\frac{8.314}{1.262-1}$

$\Rightarrow C_v=31.73J/mol-K$

Hence the estimateed value of $C_v$ is $31.72J/mol-K$.