Question

Heat is added to an ideal gas, and the gas expands. In such a process the temperature
A) must always increase
B) will remain the same if the work done equals the heat added
C) must always decrease
D) will remain the same If the change in internal energy equals the heat added.

Answer 1

Hint: The first law of thermodynamics relates the heat added to the system, the work done by the system, and the internal energy of the gas. The internal energy of the gas depends on the temperature of the gas.

Formula used: In this solution, we will use the following formula:
The first law of thermodynamics: $\Delta U = \Delta Q - W$, where $\Delta U$ is the change in internal energy, $\Delta Q$, is the heat added to the system and $W$ is the work done by the system.

Complete step by step answer:
We’ve been given that heat is added to an ideal gas and the gas expands. The first formula of thermodynamics relates the heat added to the system, the work done by the system, and the internal energy of the gas. It is given as
$\Delta U = \Delta Q - W$
When heat is added to the system, $\Delta Q$ is positive and when work is done by the gas, in which case it expands in volume, $W$ is also positive.
Now, the internal energy of the gas depends only on the temperature of the gas.
So, if the heat added to the gas and work done by the gas is the same, the change in internal energy will be zero. That means the internal energy will remain the same. Hence option (B) is correct.

Note: Option (A) is often chosen but as discussed in option (B), if the work is done and the heat added to the system are equal, the internal energy of the gas will not change. We must be careful in determining the signs of the terms in the first law of thermodynamics. Heat added to the system is considered positive and work done by the gas in expanding is also considered positive.