One mole of an ideal diatomic gas expands from volume V to 2 V isothermally at a temperature \(27^{\circ} C\) and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from \(27^{\circ} C\) doing the same amount of work W , then its final temperature will be (close to) ___________ \({ }^{\circ} C\).

For Isothermal process
\(W _{\text {isothermal }}= nRT \ell n \left(\frac{ V _2}{V_1}\right)\)
\(=1 . R 300 . \ell n (2)\)
\(=300 R(0.693)\)............(1)
Now for adiabatic process,
It is given work done in Isothermal = work done in adiabatic
\(W_{\text {adiabatic }}=\frac{nR\left(T_1-T_2\right)}{\gamma-1}\)............(2)
\((1)=(2)\)
\(\frac{ nR \left(300- T _{\text {final }}\right)}{1.4-1}=300 R (0.693)\)
\(T _{\text {Final }}=216.84 K\)
\(=-56^{\circ} C\)