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Q.
One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, $27^{\circ}C$. The work done on the gas will be :
Work done in isothermal process is given as
$W=-n R T \ln \frac{V_{2}}{V_{1}}$.
Using $P_{1} V_{1}=P_{2} V_{2}$,
we get $\frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}$.
Therefore, $W=-n R T \ln \frac{P_{1}}{P_{2}}$
Now given that $T=27{ }^{\circ} C =300\, K$
$\left(\because 0^{\circ} C =273\, K \Rightarrow 27{ }^{\circ} C \Rightarrow 27+273\, K \right)$ and $P_{2}=2 P_{1}$
Therefore, $W=-300\, n R \ln \frac{P_{1}}{2 P_{1}}$
Also, $n=1 \Rightarrow W=-300\, R \ln \left(\frac{1}{2}\right)$.
Now, using $\ln \left(\frac{a}{b}\right)=\ln (a)-\ln (b)$, we get
$W=-300\, R[\ln (1)-\ln (2)]=-300\, R[0-\ln 2]$
$\Rightarrow W=300 R \ln 2$