91Ó°ÊÓ

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 \(85\;{\rm{J}}\).