Question

Three moles of argon gas (assumed to be an ideal gas) originally at$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{Pa}$ and a volume of $\mathbf{0}\mathbf{.}\mathbf{028}{\mathbf{m}}^{\mathbf{3}}$are first heated and expanded at constant pressure to a volume of role="math" localid="1668082953575" $\mathbf{0}\mathbf{.}\mathbf{0435}{\mathbf{m}}^{\mathbf{3}}$, then heated at constant volume until the pressure reaches $\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{Pa}$, then cooled and compressed at constant pressure until the volume is again $\mathbf{0}\mathbf{.}\mathbf{028}{\mathbf{m}}^{\mathbf{3}}$, and finally cooled at constant volume until the pressure drops to its original value of $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{Pa}$. (a) Draw the pV-diagram for this cycle. (b) Calculate the total work done by (or on) the gas during the cycle. (c) Calculate the net heat exchanged with the surroundings. Does the gas gain or lose heat overall?

Short answer

(a) The p-V diagram

(b) ${\mathrm{W}}_{\mathrm{t}}=-310\mathrm{J}$

(c) $\mathrm{Q}=-310\mathrm{J}$, heat is liberated from the gas

Step-by-step solution

Draw the p-V diagram

In the first and third processes, pressure is constant and in the second and fourth processes, volume is constant. The graph will be as follows

(b) Calculate the total work done

The total work will be the sum total of the works involved in all the four steps.

${\mathrm{W}}_{\mathrm{t}}={\mathrm{W}}_1+{\mathrm{W}}_2+{\mathrm{W}}_3+{\mathrm{W}}_4$

But in the second and fourth processes, the volume is constant. So, the work done in those steps will be zero, that is ${\mathrm{W}}_2={\mathrm{W}}_4=0$

Given that for the first process,

The initial pressure, ${\mathrm{p}}_1=1.5\times {10}^4\mathrm{Pa}$

The initial volume, ${\mathrm{V}}_1=0.028{\mathrm{m}}^3$

The final volume,${\mathrm{V}}_2=0.0435{\mathrm{m}}^3$

The work for the first process:

role="math" localid="1668083550068" $$\begin{gathered} {\mathrm{W}}_1={\mathrm{P}}_1\left({\mathrm{V}}_2-{\mathrm{V}}_1\right) \\ =1.5\times {10}^4\times (0.0435-0.028) \\ =233\mathrm{J} \end{gathered}$$

Similarly, for the third process,

The final pressure,${\mathrm{p}}_2=3.5\times {10}^4\mathrm{Pa}$

The initial volume,${\mathrm{V}}_1=0.0435{\mathrm{m}}^3$

The final volume,${\mathrm{V}}_2=0.028{\mathrm{m}}^3$

The work is,

$$\begin{gathered} {\mathrm{W}}_1={\mathrm{P}}_2\left({\mathrm{V}}_2-{\mathrm{V}}_1\right) \\ =3.5\times {10}^4\times (0.028-0.0435) \\ =-543\mathrm{J} \end{gathered}$$

Therefore, the total work involved will be

$$\begin{gathered} {\mathrm{W}}_{\mathrm{t}}={\mathrm{W}}_1+{\mathrm{W}}_2+{\mathrm{W}}_3+{\mathrm{W}}_4 \\ =233+0-543+0 \\ =-310\mathrm{J} \end{gathered}$$

(c) Calculate the heat exchange

According to the first law of thermodynamics,

$\mathbf{Q}\mathbf{=}\mathbf{W}\mathbf{+}\mathbf{\Delta U}$

Here, Q is the heat, W is the total work, and is the change in internal energy.

In a cyclic process,

$\mathbf{\Delta U}\mathbf{=}\mathbf{0}$

So, the heat exchange will be

$$\begin{gathered} \mathrm{Q}=\mathrm{W} \\ =-310\mathrm{J} \end{gathered}$$

Hence, the heat will be liberated from the gas.