Chapter 20, Problem 1OQ

To determine

The process in which the work done is the highest.

Answer to Problem 1OQ

Option (b) adiabatic

Explanation of Solution

The ideal gas is compressed to half of its initial volume by isothermal, adiabatic, isobaric processes.

In PV diagram of a process, the area under the curve is equal to the work done by the process. In case of isobaric process the initial and final pressure is same. so the curve will be a straight line parallel to the volume axis.

Write the formula for the final pressure in isothermal process.

    $\begin{array}{c}{P}_f=\left(\frac{V_i}{V_f}\right)P_i\\ =\left(\frac{2V_f}{V_f}\right)P_i\\ =2P_i\end{array}$

Here, $P_f$ is the final pressure, $P_i$ is the initial pressure, $V_i$ is the initial volume, and $V_f$ is the final volume.

Write the formula for the final pressure in adiabatic process.

    $\begin{array}{c}{P}_f={\left(\frac{V_i}{V_f}\right)}^{\gamma }{P}_i\\ ={\left(\frac{2V_f}{V_f}\right)}^{\gamma }{P}_i\\ =2^{\gamma }{P}_i\end{array}$

Here, $\gamma$ is the heat capacity ratio.

The figure below shows the PV diagram of the three process.

The work done in isobaric process is the area under the curve $C_1$. The work done in isothermal process is the area under the curve $C_2$. The work done in adiabatic process is the area under the curve $C_3$.

Thus the work done is the highest in case of adiabatic process. Thus option (b) is correct.

Conclusion:

The work done is the highest in case of adiabatic process. Thus option (b) is correct.

The work done in isothermal process is less than adiabatic process. Thus option (a) is incorrect.

The work done in isobaric process is less than adiabatic process. Thus option (c) is incorrect.

The work done depends on the process. Thus option (d) is incorrect.