Question: (II) An ideal gas expands at a constant total pressure of 3.0 atm from 410 mL to 690 mL. Heat then flows out of the gas at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. Calculate (a) the total work done by the gas in the process, and (b) the total heat flow into the gas.

An isothermal process may be defined as the process in which the system's temperature stays constant. In this process, the work done is due to the variation in the net heat content in the system.

Given data:

The pressure of the ideal gas is\(P = 3.0\;{\rm{atm}}\).

The initial volume is \({V_1} = 410\;{\rm{mL}}\).

The final volume is \({V_2} = 690\;{\rm{mL}}\).

(a)

The total work done by the gas in this process can be calculated as:

\(\begin{aligned}{c}W &= P\Delta V\\W &= P\left( {{V_2} - {V_1}} \right)\\W &= \left( {3.0\;{\rm{atm}}} \right)\left( {\frac{{1.01 \times {{10}^5}\;{\rm{Pa}}}}{{1\;{\rm{atm}}}}} \right)\left( {\left\{ {\left( {690\;{\rm{mL}}} \right) - \left( {410\;{\rm{mL}}} \right)} \right\}\left( {\frac{{{{10}^{ - 6}}\;{{\rm{m}}^{\rm{3}}}}}{{1\;{\rm{mL}}}}} \right)} \right)\\W &= 84.84\;{\rm{J}} \approx {\rm{85}}\;{\rm{J}}\end{aligned}\)

Thus, the total work done by the gas in this process is \({\rm{85}}\;{\rm{J}}\).

(b)

The variation in the temperature during the entire process is constant. Therefore, the change in the internal energy will be zero. That is \(\Delta U = 0\).

The total heat flow into the gas can be calculated as:

\(\begin{aligned}{c}\Delta U &= Q - W\\0 &= Q - \left( {85\;{\rm{J}}} \right)\\Q &= 85\;{\rm{J}}\end{aligned}\)

Thus, the total heat flow into the gas is \(85\;{\rm{J}}\).