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Q.
Work done during isothermal expansion of one mole of an ideal gas from $10$ atm to $1$ atm at $300\, K$ is:
BHUBHU 2004
Solution:
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 , $
$P_{1}=10 \,atm$
$W=2.303 \,nRT \log \frac{P_{2}}{P_{1}}$
$=2.303 \times 1 \times 8.314 \times 300 \times \log \frac{1}{10}$
$=5744.1 \,J$