Chapter 4.5, Problem 1P

An ideal gas at a given state expands to a fixed final volume first at constant pressure and then at constant temperature. For which case is the work done greater?

To determine

Which case had greater work done in an ideal gas at a given state which expands to a fixed volume at constant pressure then at constant temperature.

Explanation of Solution

Refer to Figure 4-3, the boundary work which is equal to $Pd\nu$ is described under the process curve of $P-\nu$ diagram for the quasi-equilibrium process.

Write the equation of area under the process curve of $P-\nu$ diagram.

$\begin{array}{c}A={\displaystyle \underset{1}{\overset{2}{\int }}dA}\\ ={\displaystyle \underset{1}{\overset{2}{\int }}Pd\nu }\end{array}$

Here, pressure is P, change in the volume is $d\nu$, and change in the area is $dA$.

From above equation, the constant pressure case had a greater work done.