Question

0.10 mol of nitrogen gas follow the two processes shown in

Figure EX19.31. How much heat is required for each?

Short answer

The heat required for different process are -1000 j and 700 J.

Step-by-step solution

Arranging the equations

This is an isochoric process, and the heat will be given by

$Q=n{C}_v∆T$, where the temperature difference, role="math" localid="1649081855608" $∆T=T_f-T_1$

f denotes final parameters and i denotes intial parameters.

finding the needed heat

For process A:

if the temperature decreased by a factor of 3, the temperature did so as well. So, the temperature difference, $∆T=-2T_f$.

As per the ideal gas law,

$$\begin{gathered} p_f{V}_f=nR{T}_f \\ {T}_f=\frac{p_f{V}_f}{nR} \end{gathered}$$

Therefore, the heat will be

role="math" localid="1649081405766" $Q=-2n{C}_v\frac{p_f{V}_f}{nR}=\frac{p_f{V}_f{C}_v}{R}\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \left(1\right)$

The nitrogen gas has an isochoric heat capacity of 20.8J/mol$·$K.

So,

role="math" localid="1649081654971" $Q=\frac{-2·{10}^5·2000·{10}^{-6}·20.8}{8.314}=-1000J$

For process B:

with the same process of A,

$Q=\frac{2·29.1·1·{10}^5·1000·{10}^{-6}}{8.314}=700J$