Question: (II) The pressure in an ideal gas is cut in half slowly, while being kept in a container with rigid walls. In the process, 465 kJ of heat left the gas. (a) How much work was done during this process? (b) What was the change in internal energy of the gas during this process?

The constant volume process may be defined as the thermodynamic process in which the system's volume stays constant and pressure changes accordingly.

Given data:

The heat flows out of the system is \(Q = - 465\;{\rm{kJ}}\).

(a)

Since the walls of the container are rigid, therefore, there will be no change in the volume. That is \(\Delta V = 0\).

The work done during this process can be calculated as:

\(\begin{aligned}{c}W &= P\Delta V\\W &= P\left( 0 \right)\\W &= 0\end{aligned}\)

Thus, the work done during this process is \(0\).

(b)

The change in internal energy of the gas during this process can be calculated as:

\(\begin{aligned}{c}\Delta U &= Q - W\\\Delta U &= \left( { - 465\;{\rm{kJ}}} \right) - 0\\\Delta U &= - 465\;{\rm{kJ}}\end{aligned}\)

Thus, the change in internal energy of the gas during this process is \( - 465\;{\rm{kJ}}\).