Question: (II) Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.4 atm. Then the gas expands at constant pressure, from a volume of 5.9 L to 9.3 L, where the temperature reaches its original value. See Fig.15–22. Calculate (a) the total work done by the gas in the process, (b) the change in internal energy of the gas in the process, and (c) the total heat flow into or out of the gas.

In the constant volume process, the reaction's heat is equivalent to the variation in the internal energy since the value of the work done in this process is equal to zero.

(a)

In the first step from a to be, the heat is allowed to flow out of an ideal gas at constant volume. So, the change in the volume in this process will be \(\Delta {V_{{\rm{ab}}}} = 0\). Thus, the value of the work done in this process will be:

\(\begin{aligned}{c}{W_{{\rm{ab}}}} &= P\Delta {V_{{\rm{ab}}}}\\{W_{{\rm{ab}}}} &= 0\end{aligned}\)

The process b to c is a constant pressure process. In this process, the pressure is:

\(\begin{aligned}{c}P &= \left( {1.4\;{\rm{atm}}} \right)\left( {\frac{{1.01 \times {{10}^5}\;{\rm{Pa}}}}{{1\;{\rm{atm}}}}} \right)\\P &= 1.414 \times {10^5}\;{\rm{Pa}}\end{aligned}\)

The change in the volume is:

\(\begin{aligned}{c}\Delta {V_{{\rm{bc}}}} &= \left( {\left( {9.3\;{\rm{L}}} \right) - \left( {5.9\;{\rm{L}}} \right)} \right)\left( {\frac{{{{10}^{ - 3}}\;{{\rm{m}}^{\rm{3}}}}}{{1\;{\rm{L}}}}} \right)\\\Delta {V_{{\rm{bc}}}} &= 3.4 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{3}}}\end{aligned}\)

So, the work done in process b to c will be:

\(\begin{aligned}{c}{W_{{\rm{bc}}}} &= P\Delta {V_{{\rm{bc}}}}\\{W_{{\rm{bc}}}} &= \left( {1.414 \times {{10}^5}\;{\rm{Pa}}} \right)\left( {3.4 \times {{10}^{ - 3}}\;{{\rm{m}}^{\rm{3}}}} \right)\\{W_{{\rm{bc}}}} &= 480\;{\rm{J}}\end{aligned}\)

The total work done in two-step process can be calculated as:

\(\begin{aligned}{c}W &= {W_{{\rm{bc}}}} + {W_{{\rm{ab}}}}\\W &= \left( {480\;{\rm{J}}} \right) + 0\\W &= 480\;{\rm{J}}\end{aligned}\)

Thus, the total work done in two-step process is \(480\;{\rm{J}}\).

(b)

The expression for the internal energy of a monoatomic gas is as follows:

\(\Delta U = \frac{3}{2}nR\Delta T\)

In this two-step process, there is no change in the temperature. Therefore, the value of the change in the internal energy will be:

\(\Delta U = 0\)

Thus, the change in internal energy of the gas in the process is \(0\).

(c)

Thetotal heat flow into or out of the gas can be calculated as:

\(\begin{aligned}{c}\Delta U &= Q - W\\0 &= Q - \left( {480\;{\rm{J}}} \right)\\Q &= 480\;{\rm{J}}\end{aligned}\)

Since, the sign is positive, the heat flows into the system.

Thus, the heat absorbed in this two-step process is \(480\;{\rm{J}}\).