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One mole of an ideal monoatomic gas undergoes the following four reversible processes:
Step $1$ It is first compressed adiabatically from volume $8.0 \,m ^{3}$ to $1.0 \,m ^{3}$.
Step $2$ Then expanded isothermally at temperature $T_{1}$ to volume $10.0 \,m ^{3}$.
Step $3$ Then expanded adiabatically to volume $80.0 \,m ^{3}$.
Step $4$ Then compressed isothermally at temperature $T_{2}$ to volume $8.0 \,m ^{3}$.
Then, $T_{1} / T_{2}$ is
$2$
$4$
$6$
$8$
Solution
$(b)$ Given,
Step $1 A$ to $B$ adiabatic compression, $V_{A}=8 \,m ^{3}, V_{B}=1 \,m ^{3}$
Step $2 B$ to $C$ isothermal expansion,
$T_{C}=T_{B}=T_{1}, V_{C}=10 \,m ^{3}$
Step $3 C$ to $D$ adiabatic expansion,
$V_{D}=80 \,m ^{3}$
Step $4 D$ to $A$ isothermal compression,
$T_{D}=T_{A}=T_{2}, V_{A}=8 \,m ^{3}$
Now, for processes $A$ to $B$,
$T_{A} V_{A}^{\gamma-1}=T_{B} V_{B}^{\gamma-1}$
Substituting values in above equation, we get
$T_{2}(8)^{\frac{5}{3}-1}=T_{1}(1)^{\frac{5}{3}-1}$
$\therefore$ For ideal monoatomic gas, $\gamma=\frac{5}{3}$
So, $\quad \frac{T_{1}}{T_{2}}=8^{\frac{2}{3}}=4$
