Question

A sample of argon gas at $1$ atm pressure and $27{}^{\circ }C$ expands reversibly and adiabatically from $1.25d{m}^3$ to $2.50d{m}^3$. Calculate the enthalpy change in this process $C_{v.m}$ for Argon is $12.48J{K}^{-1}mo{l}^{-1}$

Answer 1

Answer: -117.25

Given : $P=1$ atm
$T_1=300K$
$V_1=1.25L$
$V_2=2.5L$
$\Delta H=$ ?
$C_{v.m.}=12.48J{K}^{-1}mo{l}^{-1}$
$\Delta H=n\times Cp\times \Delta T$
$Cp=Cv+R=12.48+8.314$
$=20.794J{K}^{-1}mo{l}^{-1}$
$=n=\frac{PV}{RT}=\frac{1\times 1.25}{0.0821\times 300}=0.05$
From $T_1{V}_1{}^{\gamma -1}=T_2{V}_2^{\gamma -1}$
$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
$=300\times {\left(\frac{1.25}{2.50}\right)}^{1.66-1}=300\times {\left(\frac{1}{2}\right)}^{0.66}$
$\therefore T_2=189.85K$
$\therefore \Delta T=T_2-T_1$
$=189.85-300$
$=-110.15k$
$\therefore \Delta H=0.05\times 20.794\times (-110.15)$
$=-114.52J$