Question

An ideal gas undergoes a reversible isothermal expansion at $\mathbf{77}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$, increasing its volume from$\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{L}$to$\mathbf{3}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{L}$. The entropy change of the gas is$\mathbf{22}\mathbf{.}\mathbf{0}\mathbf{J}\mathbf{/}\mathbf{K}$. How many moles of gas are present?

Short answer

There are 2.75 moles of gas present in the expansion.

Step-by-step solution

The given data

1. Temperature at which the isothermal expansion takers place,$T=77°C\mathrm{or}350\mathrm{K}$
2. The volume change takes place fromto$V_i=1.30\mathrm{L}\mathrm{to}{\mathrm{V}}_{\mathrm{f}}=3.40\mathrm{L}$
3. The entropy change of the gas,$∆S=22.0\mathrm{J}/\mathrm{K}$

Understanding the concept of entropy change

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system.We can write the formula for change in entropy for an isothermal process. We can rearrange the formula for the number of moles. Then inserting the given values, we can get the number of moles.

Formula:

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

Where, is gas constant = $8.314\mathrm{J}.{\mathrm{mol}}^{-1}.{\mathrm{K}}^{-1}$

Calculation of the number of moles of the gas

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

From equation (i) and the given values, we can get the number moles of the gas present during the expansion as follows:

$$\begin{gathered} ∆S=nRIn\left(\frac{V_f}{V_i}\right) \\ n=\frac{∆S}{nRIn\left(\frac{V_f}{V_i}\right)} \\ =\frac{22\mathrm{J}/\mathrm{K}}{\left(8.314\mathrm{J}.{\mathrm{mol}}^{-1}.{\mathrm{K}}^{-1}\right)\mathrm{In}\left({\displaystyle \frac{3.40\mathrm{L}}{1.30\mathrm{L}}}\right)} \end{gathered}$$

Hence, there are present 2.75 mol in the gas.