Question

Two mole of ideal gas diatomic gas ($C_{v,m}=5/2R$) at $300K$ and $5$atm expanded irreversely and adiabatically to a final pressure of $2$ atm against a constant pressure of $1$ atm. Calculate $q,w,\Delta H$ and $\Delta U$.

• A. $q=0,\Delta U=w=-1247.1J,\Delta H=-1745.94J$
• B. $q=0,\Delta U=w=+1247.1J,\Delta H=-1745.94J$
• C. $U=0,\Delta q=w=+1247.1J,\Delta H=+1745.94J$
• D. None of these

Answer 1

The correct option is A $q=0,\Delta U=w=-1247.1J,\Delta H=-1745.94J$

For adiabatic change, there is no heat transfer involved.
$q=0$

The change in the internal energy is equal to the work done $\Delta U=w$

But $\Delta U=nCv\Delta T$ and $w=P\Delta V=P_{avg}(\frac{nR{T}_f}{P_f}-\frac{nR{T}_i}{P_i})$

Hence,
$nCv\Delta T=P_{avg}(\frac{nR{T}_f}{P_f}-\frac{nR{T}_i}{P_i})$

$n\times \frac{5}{2}R\Delta T=-P_{avg}(\frac{nR{T}_f}{P_f}-\frac{nR{T}_i}{P_i})$

$5/2(T_f-300)=-(\frac{T_f}{2}-\frac{300}{5})$

$5/2T_f-750=-\frac{T_f}{2}+60$

$3T_f=810$ ; $T_f=270K$

The change in the internal energy is

$\Delta U=w=n{C}_V\Delta T=2\times \frac{5}{2}R(-30)=-150R=-1247.1J$

The enthalpy change is $\Delta H=\Delta U+nR\Delta T=-150R+2R(-30)=-210R=-1745.9J$