#
Q.34

**Q34.** 1 mole of a diatomic gas is contained in a piston. It gains 50.0J of heat and work is done on the surrounding by system is - 100 J. Thus,

$\left(A\right)thegaswillcoolby2.41\xb0\phantom{\rule{0ex}{0ex}}\left(B\right)thegaswillheatby2.41\xb0\phantom{\rule{0ex}{0ex}}\left(C\right)thegaswillcoolby3.61\xb0\phantom{\rule{0ex}{0ex}}\left(D\right)thegaswillheatby3.61\xb0$

$\u2206T\hspace{0.17em}=\frac{C}{\u2206Q}=\frac{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}R}{-100+50}=-2.41$

The correct option is (a) the gas will cool by 2.41

^{0}.

**
**