Q. One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) $ \rightarrow $ (4.0 atm, 5.0 L, 245 K) with a change in internal energy, $\Delta U$ = 30.0 L atm. The change in enthalpy $\left(\right.\Delta H\left.\right)$ of the process in L atm is
NTA AbhyasNTA Abhyas 2022
Solution:
Relation between enthalpy change and internal energy change for non-ideal gas:
$\Delta H=\Delta U+\Delta \left(\right.PV\left.\right)$
$=\Delta U+\left(\right.P_{2}V_{2}-P_{1}V_{1}\left.\right)$
$=30+\left(\right.4\times 5-2\times 3\left.\right)$
$=30+14=44$ L.atm
$\Delta H=\Delta U+\Delta \left(\right.PV\left.\right)$
$=\Delta U+\left(\right.P_{2}V_{2}-P_{1}V_{1}\left.\right)$
$=30+\left(\right.4\times 5-2\times 3\left.\right)$
$=30+14=44$ L.atm