Question

One mole of an ideal gas at $22.4\text{L}$ is expanded isothermally and reversibly at $300\text{K}$ to a volume of $224\text{L}$ . Which of the following options are correct?
(Given : $R=2{\text{cal K}}^{-1}{\text{mol}}^{-1}$)

• A. Change in enthalpy is $0$
• B. The magnitude of work done in the process is $1.38\text{kcal}$
• C. Heat change during the process is $-1.38\text{kcal}$
• D. Change in internal energy is $10\text{kcal}$

Answer 1

Answer: (A), (B)

The magnitude of work done in the process is $1.38\text{kcal}$

Here $n=1\text{mole}{V}_i=22.4\text{L}{V}_f=224\text{L}T=300\text{K}$
$W=-2.303\times nRT\log \frac{V_f}{V_i}=-2.303\times 1\times 2\times {10}^{-3}\times 300\times \log \frac{224}{22.4}=-2.303\times 2\times 0.3=-1.38\text{kcal}$
In isothermal process,
$\Delta E=0\therefore Q+W=0\Rightarrow Q=-W=1.38\text{kcal}$
As $\Delta T=0,$ then $\Delta H=n{C}_p\Delta T=0$