Question

Work done during isothermal expansion of one mole of an ideal gas from $10$ atm to $1$ atm at $300\, K$ is:

• A. 4938.8 J
• B. 4138.8 J
• C. 5744.1 J
• D. 6257.2 J

Answer 1

Answer: (C)

5744.1 J

Answer 2

For calculating work done during isothermal expansion of ideal gas we use the formula, $W=2.303 \,n R T \log \frac{P_{2}}{P_{1}}$ $W =$ work done, $n=1,$ $T=300\, K,$ $P_{2}=1 \,atm ,$ $P_{1}=10 \,atm$ $W=2.303 \,nRT \log \frac{P_{2}}{P_{1}}$ $=2.303 \times 1 \times 8.314 \times 300 \times \log \frac{1}{10}$ $=5744.1 \,J$