HeatQ flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas?

The heat energy shows the dependency on the system’s work done and the system’s internal energy. The work establishes the relation between pressure and change in volume.

The work done due to volume change at a constant pressure is given by:

$$\begin{gathered} W=p∆V \\ W=p\left(V_2-V_1\right) \end{gathered}$$

Here, p is the pressure,$v_1$ is the initial volume,$v_2$is the final volume.

The ideal gas equation is:

$p\Delta V=nR\Delta T$

Here,$∆V$is the change in the volume, n is the number of moles, R is the universal gas constant, andis the change in the temperature.

The change in internal energy is given by:

$\Delta U=n{C}_V\Delta T$

Here,$C_v$is the specific heat at constant volume.

According to the first law of thermodynamics,

$Q=W+\Delta U$

Here, W is the work done, and $∆U$is the change in the internal energy.

Use the first law and other relations to find the work.

$$\begin{gathered} Q=W+\Delta U \\ Q=p\Delta V+\Delta U \\ Q=nR\Delta T+n{C}_V\Delta T \end{gathered}$$

It can be further solved as:

$$\begin{gathered} Q=n∆T\left(R+1.5R\right) \\ Q=2.5n∆TR \\ Q=2.5p∆V \\ Q=2.5W \end{gathered}$$

Here, for an ideal gas, the term$C_V$ is equal to 1.5R.

It can be solved further as:

$$\begin{gathered} Q=\frac{5}{2}W \\ \frac{W}{Q}=\frac{2}{5} \\ W=\frac{2}{5}Q \end{gathered}$$

Hence, the work done is $\frac{2}{5}Q$.