Question

One gram mol of a diatomic gas $(\gamma =1.4)$ is compressed adiabatically so that its temperature rises from ${27}^{o}C$ to ${127}^{o}C.$ The work done will be

• A. 2078.5 joules
• B. 207.85 joules
• C. 207.85 ergs
• D. None of the above

Answer 1

The correct option is A 2078.5 joules

We know, from first law, $\Delta U=q+w$.
For adibatic process, $q=0$.
Also $\Delta U$ = $n{C}_{v}dT$ and w = $\Delta U$ = $\frac{nR}{(\gamma -1)}(T_2-T_1)$
= $\frac{1\times R}{(1.4-1)}(400-300)=2078.5J$