Two moles of an ideal gas initially at ${27}^{0}C$ and one atmospheric pressure are compressed isothermally and reversibly till the final pressure of the gas is $10atm$. Calculate $q$ for the process.

The correct option is B $-11488J$
work done in an isothermal reversible process is given as:

$w=-2.303nRT\left(\log \frac{P_1}{P_2}\right)$
Here, $n=2mol$
$T=27+273=300K$
$P_1=1atm,P_2=10atm$
$R=8.314J{K}^{-1}mo{l}^{-1}$

$w=-2.303\times 2\times 8.314\times 300\times \log \frac{1}{10}$
$w=11488J$

For isothermal compression of ideal gas, $△U$ = 0 (as temperature is constant)

Further $△U=q+w$
$q=-w=-11488J$