In a thermodynamic process, fixed mass of a gas is changed in such a manner that the gas release $20…

In a thermodynamic process, fixed mass of a gas is changed in such a manner that the gas release $20 \mathrm{~J}$ of heat and $8 \mathrm{~J}$ of work was done on the gas. If the initial internal energy of the gas was $30 \mathrm{~J}$, the final internal energy will be
  1. 2 joule
  2. 18 joule
  3. 42 joule
  4. 58 joule

Solution

According to first law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$
$\Delta \mathrm{Q}=$ heat absorbed by gas $\Delta \mathrm{W}=$ work done by gas.
$-20 \mathrm{~J}=\Delta \mathrm{U}-8 \mathrm{~J}$
$\Delta \mathrm{U}=-12 \mathrm{~J}=\mathrm{U}_{\text {Final }}-\mathrm{U}_{\text {initial }}$
$\mathrm{U}_{\text {initial }}=30 \mathrm{~J}$
$\mathrm{U}_{\text {Final }}=30-12=18 \mathrm{~J}$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more THERMODYNAMICS questions on Aicharya