Question

A quasi-static cycle of a monoatomic ideal gas contains an isothermal process \((ab)\), followed by an isochoric process \((bc)\) and an adiabatic process \((ca)\) as shown in the figure. The volumes of the gas are \(V_1\) and \(V_2\) at \(a\) and \(b\), respectively. If the cycle has heat input \(Q_{\mathrm{in}}\) and output \(Q_{\mathrm{out}}\), then the efficiency of the cycle is defined as \[ \eta=\frac{Q_{\mathrm{in}}-Q_{\mathrm{out}}}{Q_{\mathrm{in}}} \] The correct statement(s) is/are: \[ [\text{Given: }\ln2\approx0.7] \]

• A. If \(\dfrac{V_2}{V_1}=8\), the heat released in process \(bc\) is smaller than the heat absorbed in process \(ab\)
• B. For a given value of \(\dfrac{V_2}{V_1}\), \(\eta\) does not depend on the temperature of the isothermal process
• C. If \(\dfrac{V_2}{V_1}=8\), then temperature at \(a\) is \(4\) times temperature at \(c\)
• D. If \(\dfrac{V_2}{V_1}=8\), then pressure at \(a\) is \(4\) times pressure at \(b\)

Hint:

Try building one general formula for the efficiency purely in terms of the volume ratio r=V2/V1 before checking any option -- once every term shows the same factor of Ta, you will immediately see whether efficiency depends on temperature, and the same formula answers the numeric options too.

Answer 1

The Correct Option is A

Step 1: Analyze the processes.
Process: \[ ab \] is isothermal. Hence: \[ T_a=T_b \] Process: \[ bc \] is isochoric. Process: \[ ca \] is adiabatic. For monoatomic gas: \[ \gamma=\frac53 \]

Step 2: Use adiabatic relation between \(c\) and \(a\).
For adiabatic process: \[ TV^{\gamma-1}=\text{constant} \] Thus: \[ T_cV_2^{2/3}=T_aV_1^{2/3} \] \[ \frac{T_a}{T_c} = \left(\frac{V_2}{V_1}\right)^{2/3} \] If: \[ \frac{V_2}{V_1}=8 \] \[ \frac{T_a}{T_c}=8^{2/3}=4 \] Hence: \[ \boxed{\mathrm{(C)\ is\ correct}} \]

Step 3: Find pressure ratio for isothermal process.
For isothermal process: \[ PV=\text{constant} \] Thus: \[ P_aV_1=P_bV_2 \] \[ \frac{P_a}{P_b} = \frac{V_2}{V_1} \] If: \[ \frac{V_2}{V_1}=8 \] \[ \frac{P_a}{P_b}=8 \] Hence statement: \[ P_a=4P_b \] is false. Therefore: \[ \boxed{\mathrm{(D)\ is\ incorrect}} \]

Step 4: Compare heats in processes \(ab\) and \(bc\).
Heat absorbed in isothermal expansion: \[ Q_{ab}=nRT_a\ln\left(\frac{V_2}{V_1}\right) \] For: \[ \frac{V_2}{V_1}=8 \] \[ Q_{ab}=nRT_a\ln8 \] \[ =3nRT_a\ln2 \] Using: \[ \ln2\approx0.7 \] \[ Q_{ab}\approx2.1nRT_a \] Now for isochoric process: \[ Q_{bc}=nC_V(T_c-T_b) \] Since: \[ T_b=T_a,\qquad T_c=\frac{T_a}{4} \] \[ Q_{bc} = n\left(\frac32R\right)\left(\frac{T_a}{4}-T_a\right) \] \[ = -\frac98nRT_a \] Magnitude: \[ |Q_{bc}|=1.125\,nRT_a \] Thus: \[ |Q_{bc}|<Q_{ab} \] Hence: \[ \boxed{\mathrm{(A)\ is\ correct}} \]

Step 5: Check efficiency dependence.
Efficiency: \[ \eta = 1-\frac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}} \] Both heats are proportional to: \[ T_a \] Hence temperature cancels out. Thus efficiency depends only on: \[ \frac{V_2}{V_1} \] Therefore: \[ \boxed{\mathrm{(B)\ is\ correct}} \]

Step 6: Identify correct statements.
Therefore: \[ \boxed{\mathrm{(A),\ (B)\ and\ (C)}} \]

Answer 2

Concept:

• Build one general formula for the efficiency of the cycle in terms of the volume ratio $r=V_2/V_1$ first, before plugging in any specific number -- this settles the temperature-dependence question in one shot and reuses the same formula for every other check.
• For an isothermal step, $PV=\text{constant}$ gives the pressure ratio directly from the volume ratio.
• For an adiabatic step, $TV^{\gamma-1}=\text{constant}$ gives the temperature ratio directly from the volume ratio.
• Both heats in this cycle carry a common factor of $T_a$, so that temperature always cancels out of any ratio built from them, such as efficiency.

Step 1: Set up the temperature and pressure ratios in terms of $r$.
Adiabatic leg $c \to a$: $T_c V_2^{\gamma-1}=T_a V_1^{\gamma-1}$, so $\frac{T_a}{T_c}=r^{\gamma-1}$. For a monoatomic gas $\gamma=\frac{5}{3}$, so $\frac{T_a}{T_c}=r^{2/3}$.
Isothermal leg $a \to b$: $P_a V_1=P_b V_2$, so $\frac{P_a}{P_b}=r$.

Step 2: Build $Q_{in}$ and $Q_{out}$ symbolically, keeping $T_a$ as a letter, not a number.
$Q_{in}=Q_{ab}=nRT_a\ln r$.
Since $T_b=T_a$ and $T_c=T_a r^{-2/3}$, the isochoric leg gives $Q_{out}=|Q_{bc}|=nC_v(T_b-T_c)=\frac{3}{2}nRT_a\left(1-r^{-2/3}\right)$.

Step 3: Build $\eta(r)$ and watch $T_a$ cancel -- this settles statement (B).
$\eta=1-\frac{Q_{out}}{Q_{in}}=1-\frac{\frac{3}{2}nRT_a\left(1-r^{-2/3}\right)}{nRT_a\ln r}=1-\frac{3\left(1-r^{-2/3}\right)}{2\ln r}$.
$T_a$ sits in both the numerator and denominator, so it cancels completely -- efficiency depends only on $r$, for every value of $r$, not just $8$. Statement (B) is TRUE.

Step 4: Plug $r=8$ into Step 1 to settle statements (C) and (D).
$\frac{T_a}{T_c}=8^{2/3}=(2^3)^{2/3}=2^2=4$, so $T_a=4T_c$, matching statement (C). Statement (C) is TRUE.
$\frac{P_a}{P_b}=8$, not $4$, so statement (D) is FALSE.

Step 5: Plug $r=8$ into Step 2 to settle statement (A).
$Q_{in}=nRT_a\ln 8=3nRT_a\ln2\approx3(0.7)nRT_a=2.1\,nRT_a$.
$Q_{out}=\frac{3}{2}nRT_a\left(1-8^{-2/3}\right)=\frac{3}{2}nRT_a\left(1-\frac{1}{4}\right)=\frac{3}{2}nRT_a\left(\frac{3}{4}\right)=1.125\,nRT_a$.
Since $1.125\,nRT_a$ is smaller than $2.1\,nRT_a$, the heat released in $bc$ is indeed smaller than the heat absorbed in $ab$. Statement (A) is TRUE.

Final Answer: Statements (A), (B) and (C) are correct; (D) is incorrect.