Question

If $1.0kcal$ of heat is added to $1.2L$ of $O_2$ in a cylinder at constant pressure of $1$ atm, the volume increase to $1.5L$. Calculate $\Delta U$ and $\Delta H$ of the process $(1L-atm=100J)$

Answer 1

As we know that,

$\Delta U=q-P\Delta V$

Here, $q=1\text{kcal}=1000\text{cal.}$

$P=\text{Pressure}=1\text{atm}$

$\Delta V=$ Change in volume $=(1.5-1.2)L=0.3L$

$\therefore \Delta U=1000-1\times 0.3=992.7\text{cal.}=0.993\text{kcal}$

At constant pressure,

$\Delta H=q_P=1\text{kcal}$

Hence, $\Delta U=0.993\text{kcal}and\Delta H=1\text{kcal}$.