Question

A 2.50molsample of an ideal gas expands reversibly and isothermally atuntil its volume is doubled. What is the increase in entropy of the gas?

Short answer

The increase in the entropy of the gas is 14.4J/K

Step-by-step solution

The given data

1. Number of moles present in the sample, n = 2.50mol
2. Temperature at which isothermal change takes place, T = 360K
3. The final volume with respect to initial volume,$v_f=2V_i$

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. Then inserting the given values, we can find the increase in entropy of the gas.

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 increase in the entropy of the gas

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

Using equation (i) and the given values, the entropy change of the gas can be given as:

$$\begin{gathered} ∆\mathrm{S}=\mathrm{nRIn}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right) \\ =\left(2.50\mathrm{mol}\right)\left(8.314\mathrm{J}.{\mathrm{mol}}^{-1}.{\mathrm{K}}^{-1}\right)\mathrm{In}\left(\frac{2{\mathrm{V}}_{\mathrm{i}}}{{\mathrm{V}}_{\mathrm{i}}}\right) \\ =14.4\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the increased entropy change of the gas is 14.4 J/K