91Ó°ÊÓ

a) The entropy change of the gas sample,$∆S_s=64\mathrm{J}/\mathrm{K}$

b) The graph of entropy change $∆S$and volume $V_f$is given.

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system. We can write the expression for the number of moles by rearranging the formula for entropy change. Then by inserting the values obtained from the given graph, we can find the number of moles in the gas sample.

Formulae:

The entropy change of the gas, $∆S=nRIn\left(\frac{V_f}{V_i}\right)-n{C}_{v}In\left(\frac{T_f}{T_i}\right)$ …(¾±)

In an isothermal process, the temperature value remains constant. So,$T_f=T_i$

Thus, using equation (i), we can get the number of moles present in the gas sample is given as follows:

(From the graph, we can infer that,$∆S=32\frac{\mathrm{J}}{\mathrm{K}},V_f=1.2{\mathrm{m}}^3,{\mathrm{V}}_{\mathrm{i}}=0.4{\mathrm{m}}^3.$)

$$\begin{gathered} n=\frac{∆S}{RIn\left({\displaystyle \frac{V_f}{V_i}}\right)} \\ =\frac{32\mathrm{J}/\mathrm{K}}{\left(8.314\mathrm{J}/\mathrm{kg}\mathrm{K}\right)In\left({\displaystyle \frac{1.2{\mathrm{m}}^3}{0.4{\mathrm{m}}^3}}\right)} \\ =3.5\mathrm{mol} \end{gathered}$$

Hence, there are 3.5mol present in the gas sample.