Question

Two moles of an ideal monatomic gas go through the cycle abc. For the complete cycle, 800 J of heat flows out of the gas. Process ab is at constant pressure, and process bc is at constant volume. States a and b have temperatures ${\mathbf{T}}_{\mathbf{a}}\mathbf{=}\mathbf{200}\mathbf{}\mathbf{K}$and ${\mathbf{T}}_{\mathbf{b}}\mathbf{=}\mathbf{300}\mathbf{}\mathbf{K}$. (a) Sketch the p-V-diagram for the cycle. (b) What is the work W for the process ca?

Short answer

The p-V diagram:

The work for the process ca is -2463 J.

Step-by-step solution

Draw the p-V diagram

In the given process, the step ab is at constant pressure, so the line will be parallel to the volume axis. In the bc step, the volume is constant, so the line will be perpendicular to the volume axis.

The process is a closed cycle, which means that $∆\mathrm{U}=0$

So, the work done will be $\mathrm{W}=\mathrm{Q}=-800\mathrm{J}$.

As the work done is negative, the gas goes under compression. The graph will be

Calculate the work for ca step

The total work involved in the whole process is ${\mathrm{W}}_{\mathrm{t}}={\mathrm{W}}_{\mathrm{ab}}+{\mathrm{W}}_{\mathrm{bc}}+{\mathrm{W}}_{\mathrm{ca}}$.

Where, ${\mathrm{W}}_{\mathrm{bc}}=\mathrm{p}∆\mathrm{V}$as the volume is kept constant. For ab, as the pressure is kept constant, ${\mathrm{W}}_{\mathrm{ab}}=\mathrm{p}∆\mathrm{V}$. So, we can write

$$\begin{gathered} {\mathrm{W}}_{\mathrm{ca}}={\mathrm{W}}_{\mathrm{t}}-\left({\mathrm{W}}_{\mathrm{ab}}+{\mathrm{W}}_{\mathrm{bc}}\right) \\ =-800-\left(\mathrm{p}∆\mathrm{V}+0\mathrm{\pi }\right) \\ =-800-\mathrm{p}∆\mathrm{V} \end{gathered}$$

From the ideal gas equation,

$\mathbf{p}\mathbf{∆}\mathbf{V}\mathbf{=}\mathbf{nR}\mathbf{∆}\mathbf{T}$

The number of moles, n = 2moles ,

Universal gas constant, R = 8.314 J/mol

Change in temperature, $∆\mathrm{T}={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{a}}$. Now,

Temperature, ${\mathrm{T}}_{\mathrm{a}}=200\mathrm{K}$

Temperature, ${\mathrm{T}}_{\mathrm{b}}=300\mathrm{K}$

The work is,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{ca}}=-800-\mathrm{p}∆\mathrm{V} \\ =-800-\mathrm{nR}\left({\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{a}}\right) \\ =-800-\left[2\times 8.314\times \left(300-200\right)\right] \\ =-2463\mathrm{J} \end{gathered}$$

Hence, the work for the process ca is -2463 J.