A quasi-static cycle of a monoatomic ideal gas contains an isothermal process (a → b), followed by an isochoric process (b → c) and an adiabatic process (c → a) as shown in the figure. The volumes of the gas are V₁ and V₂ at a and b, respectively. If the cycle has heat input Q_in and output Q_out, then the efficiency of the cycle is defined as η = $\frac{Q_{\text{in}}-Q_{\text{out}}}{Q_{\text{in}}}$. The correct statement(s) is/are:

[Given: ln 2 ≈ 0.7]

The correct statements are:

• If $\frac{V_2}{V_1}=8$, the heat released in process b → c is smaller than the heat absorbed in process a → b

• For a given value of $\frac{V_2}{V_1}$, η does not depend on the temperature of the isothermal process

• If $\frac{V_2}{V_1}=8$, then temperature at a is 4 times temperature at c

Step-by-Step Analysis:

1. Understanding the processes in the cyclic diagram:

From the P-V diagram given in the image, we can identify the following processes for a monoatomic ideal gas ($\gamma =\frac{5}{3}$):

• Process a → b: Isothermal expansion at temperature $T_a=T_b=T$ from volume $V_1$ to $V_2$.

• Process b → c: Isochoric cooling at constant volume $V_b=V_c=V_2$ from temperature $T_b$ to $T_c$.

• Process c → a: Adiabatic compression from volume $V_2$ at temperature $T_c$ back to volume $V_1$ at temperature $T_a$.

2. Temperature relation for Adiabatic Process (c → a):

For an adiabatic process of an ideal gas:

$T\cdot V^{\gamma -1}=\text{constant}$

Since gas is monoatomic, $\gamma =\frac{5}{3}$, which gives $\gamma -1=\frac{2}{3}$.

Applying this between states c and a:

$T_c{V}_2^{2/3}=T_a{V}_1^{2/3}$

$\Rightarrow \frac{T_a}{T_c}={\left(\frac{V_2}{V_1}\right)}^{2/3}$

Given $\frac{V_2}{V_1}=8$:

$\frac{T_a}{T_c}=8^{2/3}={\left(2^3\right)}^{2/3}=2^2=4$

$\Rightarrow T_a=4T_c$

Thus, the temperature at a is indeed 4 times the temperature at c. This validates the third statement.

3. Comparing Heat absorbed in (a → b) and Heat released in (b → c):

Heat absorbed in isothermal process a → b:

$Q_{\text{in}}=Q_{a\to b}=nR{T}_a\ln \left(\frac{V_2}{V_1}\right)$

For $\frac{V_2}{V_1}=8$ and using $\ln 2\approx 0.7$:

$Q_{\text{in}}=nR{T}_a\ln \left(2^3\right)=3nR{T}_a\ln 2\approx 3\times 0.7nR{T}_a=2.1nR{T}_a$

Heat released in isochoric process b → c:

$Q_{\text{out}}=\left|Q_{b\to c}\right|=n{C}_v(T_b-T_c)$

Since $T_b=T_a$ and $T_c=\frac{T_a}{4}$, and for a monoatomic gas $C_v=\frac{3}{2}R$:

$Q_{\text{out}}=n\left(\frac{3}{2}R\right)\left(T_a-\frac{T_a}{4}\right)=\frac{3}{2}nR{T}_a\left(\frac{3}{4}\right)=\frac{9}{8}nR{T}_a=1.125nR{T}_a$

Comparing $Q_{\text{out}}$ and $Q_{\text{in}}$:

$1.125nR{T}_a<2.1nR{T}_a$

Thus, the heat released in process b → c is indeed smaller than the heat absorbed in process a → b. This validates the first statement.

4. Efficiency of the cycle (η):

The efficiency of the cycle is defined as:

$\eta =\frac{Q_{\text{in}}-Q_{\text{out}}}{Q_{\text{in}}}=1-\frac{Q_{\text{out}}}{Q_{\text{in}}}$

Substituting the general expressions:

$Q_{\text{in}}=nR{T}_a\ln \left(\frac{V_2}{V_1}\right)$

$Q_{\text{out}}=n{C}_v({\mathrm{Taccess}}_a-T_c)=n\left(\frac{3}{2}R\right)T_a\left[1-{\left(\frac{V_1}{V_2}\right)}^{2/3}\right]$

Taking the ratio $\frac{Q_{\text{out}}}{Q_{\text{in}}}$:

$\frac{Q_{\text{out}}}{Q_{\text{in}}}=\frac{\frac{3}{2}\left[1-{\left(\frac{V_1}{V_2}\right)}^{2/3}\right]}{\ln \left(\frac{V_2}{V_1}\right)}$

Since the temperature $T_a$ cancels out completely from the ratio, for any given value of $\frac{V_2}{V_1}$, the efficiency $\eta$ depends solely on the volume ratio and is completely independent of the temperature of the isothermal process. This validates the second statement.

Conclusion:

The three correct statements are:

1. If $\frac{V_2}{V_1}=8$, the heat released in process b → c is smaller than the heat absorbed in process a → b.

2. For a given value of $\frac{V_2}{V_1}$, η does not depend on the temperature of the isothermal process.

3. If $\frac{V_2}{V_1}=8$, then temperature at a is 4 times temperature at c.