One mole of a non-ideal gas undergoes a change of state $(2.0 \mathrm{~atm}, 3.0 \mathrm{~L}, 95…

One mole of a non-ideal gas undergoes a change of state $(2.0 \mathrm{~atm}, 3.0 \mathrm{~L}, 95 \mathrm{~K}) ightarrow(4.0 \mathrm{~atm}$,
5.0 L, $245 \mathrm{~K}$ ) with a change in internal energy, $\Delta \mathrm{U}=30.0 \mathrm{~L} \mathrm{~atm} .$ The change in enthalpy $\Delta \mathrm{H}$
of the process in $\mathrm{L}$ atm is.
  1. $40.0$
  2. $42.3$
  3. $44.0$
  4. Not defined because pressure is not constant

Solution

$\Delta \mathrm{H}=\mathrm{H}_{2}-\mathrm{H}_{1}=\left(\mathrm{E}_{2}+\mathrm{P}_{2} \mathrm{~V}_{2}ight)-\left(\mathrm{E}_{1}+\mathrm{P}_{1} \mathrm{~V}_{1}ight)$
$=\left(\mathrm{E}_{2}-\mathrm{E}_{1}ight)+\left(\mathrm{P}_{2} \mathrm{~V}_{2}-\mathrm{P}_{1} \mathrm{~V}_{1}ight)$
$=30+(4 \times 5-2 \times 3)=44 \mathrm{~L}$ atm

Asked in: JEE-TOPICTESTS-CHEMISTRY

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