Question

In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas molecules give out $30\, J$ of heat and $10 \,J$ of work is done on the gas. If the initial internal energy of the gas was $40\, J$, then the final internal energy will be:

• A. 30 J
• B. 20 J
• C. 60 J
• D. 40 J

Answer 1

Answer: (B)

Energy remains conserved in the process according to first law of thermodynamics.
If an amount of heat $Q$ is given to a system, a part of it is used in increasing the internal energy $(\Delta U)$ of the system and the rest in doing work $(W)$ by the system.
Hence, we have $Q=\Delta U+W$
(first law of thermodynamics)
Given, $Q=-30 \,J, W=-10 \,J$
$\therefore \Delta U=-30+10=-20\, J$
$\Rightarrow U_{f}=-U_{i}=-20\, J$
$\Rightarrow U_{f}-40=-20$
$\Rightarrow U_{f}=40-20=20\, J$