Question

Calculate the work done when $10$ moles of ideal gas expand isothermally and reversibly from a pressure of $10\text{atm}$ to $1\text{atm}$ at $300\text{K}$.
(Given : $R=2{\text{cal K}}^{-1}{\text{mol}}^{-1}$)

• A. $-13.82\text{kcal}$
• B. $13.82\text{kcal}$
• C. $-6\text{kcal}$
• D. $6\text{kcal}$

Answer 1

The correct option is A $-13.82\text{kcal}$

Here initial pressure $\left(P_1\right)=10\text{atm}$
Final pressure $P_2=1\text{atm}$
We know that, for an isothermal reversible process,
$\text{The work done,}w=-nRT\ln \frac{P_1}{P_2}=-2.303\times 10\times 2\times 300\times \log \frac{10}{1}=-13818\text{cal}=-13.82\text{kcal}$