Question

A gas $(C_{v,m}=\frac{5}{2}R)$ behaving ideally was allowed to expand reversibly and adiabatically from 1 litre to 32 litre. Its initial temperature was ${227}^{o}C.$ The molar enthalpy change (in $J/mol$) for the process is:

• A. $-1312.5R$
• B. $-675.5R$
• C. $-1575.8R$
• D. None of these

Answer 1

The correct option is A $-1312.5R$

For adiabatic reversible process we have :
$\frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$......eq(1)
Given : $V_1=1L$
$V_2=32L$
$T_1=227+273=500K$
$C_{v,m}=\frac{5}{2}R$
and we know $\gamma =\frac{C_{p,m}}{C_{v,m}}$
and $C_p-C_v=R$ (for ideal gas)
$C_p=C_v+R=\frac{7}{2}R$
hence, $\gamma =\frac{7}{5}$
So, putting these in eq.(1)
$T_2=T_1{\left(\frac{1}{32}\right)}^{\frac{7}{5}-1}=500{\left(\frac{1}{2^5}\right)}^{\frac{2}{5}}$
$=500(0.5{)}^2=125K$

Enthalpy change is given as
$△H=n{C}_p△T$
so, molar enthalpy change will be
$△H_m\left(J/mol\right)=C_p△T$
$△H_m=\frac{7}{2}R\times (125-500)=-1312.5R$
hence, The molar enthalpy change for the process is $-1312.5R$