Question

A sample of argon gas at $1$ atm pressure and $27^{\circ} C$ expands reversibly and adiabatically from $1.25 \,dm ^{3}$ to $2.50 \, dm ^{3}$. Calculate the enthalpy changc in this process $C_{V_{m}}$ for argon is $12.49 \,JK ^{-1} \, mol ^{-1}$.

Answer 1

Given : $ C_{V}=12.49 \Rightarrow C_{p}=20.8 $
$\Rightarrow \frac{C_{p}}{C_{V}}=\gamma=1.66$
In case of reversible adiabatic expansion :
$T V^{\gamma-1}=$ constant
$\Rightarrow \frac{T_{2}}{T_{1}}=\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=\left(\frac{V_{1}}{V_{2}}\right)^{0.66}$
$\Rightarrow T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{0.66}$
$\Rightarrow \Delta H=n C_{p} \Delta T$
$=\frac{1 \times 1.25}{0.082 \times 300} \times 20.8 \times(189.86-300) J$
$=-116.4\, J$