Question

One mole of a liquid ($1bar,100ml$) is taken in an adiabatic container and the pressure increases steeply to $100bar$. Then at constant pressure of $100bar$, volume decrases by $1mL$. Calculate $\Delta H$.

• A. $500J$
• B. $990J$
• C. $650J$
• D. $720J$

Answer 1

The correct option is B $990J$

From first law of thermodynamics,
$\Delta U=q+w$
But as the process is carried out under adiabatic condition, $q=0$.
Hence $\Delta U=w$
Further, at constant pressure of $100bar$, volume has decreased by $1mL$, therefore, work of contraction will be :
$w=P\Delta V=100bar\times 1mL$
$=(100\times {10}^{5}N{m}^{-2})\left({10}^{-6}{m}^3\right)=10J$

Hence, $\Delta U=10J$
Enthapy change is given as:
$\Delta H=\Delta U+\Delta \left(PV\right)$
$\Delta \left(PV\right)=(P_2{V}_2-P_1{V}_1)$

$P_1=1bar,V_1=100mL$
$P_2=100bar,V_2=99mL$

$\Delta \left(PV\right)=(100\times 99)-(1\times 100)$
$\Delta \left(PV\right)=9800barmL=980J$
as $1Lbar=101.3J$
so,
$\Delta H=10J+980J=990J$