Question

$1\text{mole}$ of a monoatomic ideal gas is taken along the cycle $\text{ABCA}$ as shown in the diagram $($ $P$ is pressure and $V$ is volume).

• A. Net heat absorbed by the gas in the given cycle is $\frac{PV}{2}$
• B. Net heat released by the gas in the given cycle is $\frac{PV}{2}$
• C. Ratio of specific heats in the process $\text{CA}$ to process $\text{BC}$ is $\frac{5}{3}$
• D. Ratio of specific heats in the process $\text{CA}$to process $\text{BC}$ is $\frac{7}{5}$

Answer 1

Answer: (A), (C)

Ratio of specific heats in the process $\text{CA}$ to process $\text{BC}$ is $\frac{5}{3}$

We know that in a cyclic process, change in internal energy,
$\Delta u=0.......\left(1\right)$

From, the first law of thermodynamics, we have,
$\Delta Q=\Delta u+\Delta w......\left(2\right)$

Where, $\Delta Q$ is heat supplied or released in a process.
$\Delta w$ is net work done during the process.

From $\left(1\right)$ and $\left(2\right)$,

$\Delta Q=\Delta w=$ Area under $\text{P-V}$ curve

$\Rightarrow \Delta Q=+\frac{1}{2}\times (2V-V)\times (2P-P)$ $($clockwise$)$

$\Rightarrow \Delta Q=+\frac{PV}{2}$ = Heat is absorbed

Also, we know that specific heat at constant volume,

$C_v=\frac{fR}{2}$

Where, $f$ is degrees of freedom of the gas molecule.
For monoatomic gas, $f=3$.
From above equation,

$\therefore C_v=\frac{3R}{2}$

We know that, $C_p-C_v=R$
where,
$C_p$ is specific heat at constant pressure,

$\Rightarrow C_P=R+C_v$

$=R+\frac{3R}{2}$

$\therefore C_p=\frac{5R}{2}$

Now, ratio of specific heats,
$\gamma =\frac{c_P}{c_v}$

$\Rightarrow \gamma =\frac{\frac{5R}{2}}{\frac{3R}{2}}$

$\therefore \gamma =\frac{5}{3}$

Hence, options $\left(a\right)$ and $\left(c\right)$ are correct answers.