Question

$15\text{J}$ of heat is added to a monoatomic ideal gas in a process in which the gas performs a work of $5\text{J}$ on its surroundings . Find the molar heat capacity of the gas during the process.

• A. $2R$
• B. $\frac{4}{3}R$
• C. $\frac{7}{5}R$
• D. $\frac{9}{4}R$

Answer 1

The correct option is D $\frac{9}{4}R$

Given,
Heat added to the gas $Q=+15\text{J}$
Work done by the gas $W=+5\text{J}$
According to first law of thermodynamics
$Q=\Delta U+W$
$\Rightarrow 15=\Delta U+5$
$\Rightarrow \Delta U=+10\text{J}$
But we know that, $\Delta U=n{C}_V\Delta T$
$\Rightarrow n{C}_V\Delta T=+10\text{J}....\left(1\right)$

Also, heat absorbed in the process $Q=nC\Delta T$
where $C$ is the molar heat capacity for the gas during the process.
Thus, $nC\Delta T=+15\text{J}.....\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$ we get,
$\frac{C}{C_V}=\frac{3}{2}\Rightarrow C=\frac{3C_V}{2}$
For a monoatomic gas, we know that $C_V=\frac{3R}{2}$
$\Rightarrow C=\frac{3}{2}\times \frac{3R}{2}$
$\Rightarrow C=\frac{9}{4}R$
Thus, option (d) is the correct answer.