Question

One mole of an ideal monatomic gas undergoes the following four reversible processes:
Step $1$: It is first compressed adiabatically from volume $V_1$ to $1m^3$.
Step $2$: then expanded isothermally to volume $10m^3$.
Step $3$: then expanded adiabatically to volume $V_3$.
Step $4$: then compressed isothermally to volume $V_1$.
If the efficiency of the above cycle is $3/4$ then $V_1$ is,

• A. $2m^3$ .
• B. $4m^3$ .
• C. $6m^3$ .
• D. $8m^3$ .

Answer 1

The correct option is D $8m^3$ .

Here, $V_1=V_1$

$V_2=1m^3$

$V_3=10m^3$

$V_4=V_4$

Since, efficiency $\eta =1-\frac{T_2}{T_1}$

$\therefore \frac{T_2}{T_1}=1-\eta$

$\frac{T_2}{T_1}=1-\frac{3}{4}=\frac{1}{4}$

For the path, $1\to 2$ (adiabatic process)

$\frac{T_1}{T_2}=(\frac{V_1}{V_2}{)}^{\gamma -1}$ (for monoatomic $\gamma =\frac{5}{3}$)

$4=(V_1{)}^{\frac{2}{3}}$

$V_1=8m^3$