Question

A gas $(C_{v,m}=\frac{5}{2}R)$ behaving ideally was allowed to expand reversibly and adiabatically from $1litre$ to $32litre$. It's initial temperature was ${327}^{o}C.$ The molar enthalpy change (in $Jmol{e}^{-1}$) for the process is

• A. $-1125R$
• B. $-575R$
• C. $-1575R$
• D. None of these

Answer 1

The correct option is C $-1575R$

As the gas is explanding reversibly and adiabatically, we can use the relation :
$\frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
We know that,
$C_p=C_v+R$
and
$\gamma =\frac{C_p}{C_v}=\frac{C_v+R}{C_v}$
Hence,
$\gamma =\frac{7}{5}$
$\Rightarrow T_2=T_1{\left(\frac{1}{32}\right)}^{\frac{7}{5}-1}=600\cdot {\left(\frac{1}{2^5}\right)}^{\frac{2}{5}}=600(0.5{)}^2=150K$
$△H_m=\frac{7}{2}R\times (150-600)=-1575R$