Question

An ideal monoatomic gas undergoes a process in which its internal energy U and density $\rho$ vary as $U\rho =constant.$ The ratio of change in internal energy and the work done by the gas is

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{3}{2}$

As we know that internal energy(U) of a monoatomic gas is given as $\frac{m{v}^2}{2}$
and work done (W)=F.S
where $F=\frac{mn{v}^2}{3Ln}$
and $S=L$
so $\frac{internalenergy}{workdone}$=$\frac{3}{2}$