Question

The internal energy of an ideal gas decreases by the same amount as the work done by the system.
(A) The process must be adiabatic
(B) The process must be isobaric.
(C) The temperature must decrease.
(D) The process must be isothermal.

Answer 1

Hint
The change in the internal energy of any system is given by the sum of the work done and the heat exchange occurring in the system. So when the work done is equal to change in internal energy then there is no heat exchange.

Complete step by step answer
We can find the solution to this problem from the first law of thermodynamics. The mathematical form of the first law is given by:
$\Delta U = Q + W$,
Where $\Delta U$ gives the change in the internal energy of the ideal gas $Q$ is used to denote the exchange of heat occurring in the system and $W$ gives the work done by the system.
This means that the change in internal energy and work done are related by the formula where the sum of the heat released or absorbed and the work done by the system gives the change in internal energy.
Now in this question, we are given that the amount by which the internal energy decreases is the same amount by which the work is done by the system.
Hence, there is no exchange of heat occurring in the process.
The process where no exchange of heat takes place is called adiabatic process. In this process since the internal energy of the system decreases and it is an ideal gas, so the process must be adiabatic.
Thus when the internal energy of an ideal gas decreases by the same amount as the work done by the system, then that process must be adiabatic and the temperature must decrease.
Therefore, the correct options will be both (A) and (C).

Note
From the question, we see that the work done by the system is positive. So the volume and the pressure of the system are varying. Therefore, this proves that the process cannot be isobaric. And since the temperature also decreases, the process isn’t isothermal.