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 change in this process $C_{v. m}$ for Argon is $12.48\,JK^{-1} mol^{-1}$

Answer 1

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