Question

A monatomic gas follows the process 1S 2S 3 shown in

Figure EX19.32. How much heat is needed for (a) process 1S 2

and (b) process 2S 3?

Short answer

The heat needed for process (a) is 150 J and for process (b) -90 J.

Step-by-step solution

Discussion

The heat needed for an isochoric/isobaric process is

$Q_{\raisebox{1ex}{$p$}\!\left/ \!\raisebox{-1ex}{$v$}\right.}=n{C}_{\raisebox{1ex}{$p$}\!\left/ \!\raisebox{-1ex}{$v$}\right.}∆T$

Process $1\to 2$is isobaric. Therefore, if the volume is tripled, the temperature will also be tripled. $\frac{V}{T}$remains constant in an isobaric process. The change in temperature is $∆T_{1\to 2}=2T_1$.

As the process $2\to 3$is a process going back to same isotherm,

$∆T_{2\to 3}=-2T_1$

Finding of heat required

From the ideal gas law,

$$\begin{gathered} pV=nRT \\ n=\frac{pV}{RT} \end{gathered}$$

In state 1,

$n=\frac{p_1{V}_1}{R{T}_1}$

Then, the heat needed for each isobaric process,

$Q_p=n{C}_p∆T=\frac{p_1{V}_1}{R{T}_1}·C_p·2T_1=\frac{2p_1{V}_1{C}_p}{R}$

Putting the numeric values,

$Q_p=\frac{2·3·{10}^5·100·{10}^{-6}·20.8}{8.314}=150J$

This result in an isochoric process will be

$Q_v=-\frac{Q_v}{\gamma }=\frac{150}{1.67}=-90J$

The heat at the beginning will be

role="math" localid="1649083337274" $$\begin{gathered} Q_v=n{C}_v∆T \\ {Q}_v=\frac{p_1{V}_1}{R{T}_1}·C_v·(-2T_1)=\frac{-2p_1{V}_1{C}_v}{R}=\frac{-2·3·{10}^5·100·{10}^{-4}·12.5}{8.314}=-90J \end{gathered}$$