Question

Volume of a monoatomic gas varies with temperature as shown in the figure.

Ratio of heat taken by gas and work done by gas is

• A. $\frac{5}{2}$
• B. $\frac{7}{2}$
• C. $\frac{5}{3}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{5}{2}$

The ratio of $V$ and $T$ is constant. So it is an isobaric process.
Since it is an isobaric process,
$\Delta Q=n{C}_P\Delta T$
W.D $=nR\Delta T$
so
$\frac{\Delta Q}{W.D}=\frac{\mathrm{n/}{C}_P\mathrm{\Delta /}T}{\mathrm{n/}R\mathrm{\Delta /}T}=\frac{\frac{5}{2}\mathrm{R/}}{\mathrm{R/}}\frac{\Delta Q}{W.D}=\frac{5}{2}$