Question

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

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

Answer 1

Answer: (C)

For calculating work done during isothermal expansion of ideal gas we use the formula,
$W=2.303nRT\log \frac{P_2}{P_1}$
$W=$ work done, $n=1,$
$T=300K,$
$P_2=1atm,$
$P_1=10atm$
$W=2.303nRT\log \frac{P_2}{P_1}$
$=2.303\times 1\times 8.314\times 300\times \log \frac{1}{10}$
$=5744.1J$