One mole of a monoatomic ideal gas is taken through the cycle shown in figure
A - B : adiabatic exp

Q: Why do we use $\gamma = 5/3$?

The problem specifies that the gas is monoatomic. For a monoatomic ideal gas, the degrees of freedom $f = 3$, so $\gamma = 1 + 2/f = 1 + 2/3 = 5/3$.

Q: Why is the work done in $A \rightarrow B$ positive?

During the process $A \rightarrow B$, the gas undergoes adiabatic expansion. Since the volume increases, the gas does work on the surroundings, making the work done positive.

Q: How did we know the volume is constant in $B \rightarrow C$?

The problem explicitly states that $B \rightarrow C$ is "cooling at constant volume", which means it is an isochoric process. In a $p-V$ diagram, this is represented by a vertical line.

Q: Why is the heat calculated in $B \rightarrow C$ negative?

The temperature decreases from $850 \text{ K}$ to $425 \text{ K}$. Since $\Delta T$ is negative, the heat $Q = nC_V\Delta T$ is also negative, indicating that heat is released or lost by the system.

Q: Can we find $T_D$ without the shortcut?

Yes! You could find $V_A$ and $V_B$ using the ideal gas law, then use the adiabatic relation $T_C V_C^{\gamma-1} = T_D V_D^{\gamma-1}$ to find $T_D$. However, the shortcut ratio method is much faster and less prone to calculation errors.

Question

One mole of a monoatomic ideal gas is taken through the cycle shown in figure
$A ightarrow B$ : adiabatic expansion
$B ightarrow C$ : cooling at constant volume
$C ightarrow D$ : adiabatic compression
$D ightarrow A$ : heating at constant volume.
The pressure and temperature at $A$, $B$, etc., are denoted by $p_A$, $T_A$, $p_B$, $T_B$ etc., respectively. Given that, $T_A = 1000 	ext{ K}$, $p_B = (2/3)p_A$ and $p_C = (1/3)p_A$, calculate the following quantities
(a) The work done by the gas in the process $A ightarrow B$.
(b) The heat lost by the gas in the process $B ightarrow C$.
(c) The temperature $T_D$. (Given : $(2/3)^{2/5} = 0.85$)

Sigma Percentile · Accepted Answer

The study of thermodynamic cycles is like deciphering the heartbeat of an engine. In this classic JEE problem, we are presented with a fascinating cycle consisting of two adiabatic and two isochoric processes. Let's break down the physics and the math step-by-step to uncover the secrets of this monoatomic ideal gas.

### Analyzing the Setup

We are given one mole of a monoatomic ideal gas. The moment you see "monoatomic", your mind should immediately register the adiabatic index, $\gamma = \frac{5}{3}$. The cycle starts at state $A$ with a high temperature $T_A = 1000 	ext{ K}$. The gas then undergoes an adiabatic expansion to state $B$, where the pressure drops to $p_B = \frac{2}{3}p_A$.

This is followed by isochoric (constant volume) cooling to state $C$, where the pressure further drops to $p_C = \frac{1}{3}p_A$. The cycle is completed by an adiabatic compression to state $D$ and finally isochoric heating back to state $A$.

### The Master Equation for Adiabatic Expansion

Our first objective is to find the work done during the adiabatic expansion from $A$ to $B$. To do this, we need the temperature at state $B$. Since the process is adiabatic, there is no heat exchange, and the pressure and temperature are linked by the beautiful relation:

$$p_A^{1-\gamma} T_A^\gamma = p_B^{1-\gamma} T_B^\gamma$$

Rearranging this to solve for $T_B$, we get:

$$T_B = T_A \left( \frac{p_A}{p_B} ight)^{\frac{1-\gamma}{\gamma}}$$

Substituting the given values, the pressure ratio $\frac{p_A}{p_B}$ is simply $\frac{3}{2}$. The exponent evaluates to $\frac{1 - 5/3}{5/3} = -\frac{2}{5}$. Flipping the fraction changes the sign of the exponent, giving us:

$$T_B = 1000 \left( \frac{2}{3} ight)^{2/5}$$

The problem kindly provides the value of $\left( \frac{2}{3} ight)^{2/5} = 0.85$. Multiplying this by $1000$ gives us $T_B = 850 	ext{ K}$.

### Calculating the Work Done

With the temperatures at both states known, calculating the work done is a breeze. The work done in an adiabatic process is given by:

$$W_{AB} = \frac{nR(T_A - T_B)}{\gamma - 1}$$

Plugging in $n = 1$, $R = 8.31 	ext{ J/mol}\cdot	ext{K}$, and our temperatures:

$$W_{AB} = \frac{1 	imes 8.31 	imes (1000 - 850)}{\frac{5}{3} - 1} = \frac{8.31 	imes 150}{2/3} = 1869.75 	ext{ J}$$

This positive value confirms that the gas is expanding and doing work on its surroundings.

### Isochoric Cooling and Heat Loss

Next, the gas cools at a constant volume from $B$ to $C$. In an isochoric process, Gay-Lussac's Law tells us that pressure is directly proportional to temperature ($p \propto T$).

Since the pressure drops from $\frac{2}{3}p_A$ to $\frac{1}{3}p_A$, it is exactly halved. Therefore, the temperature must also halve:

$$T_C = \frac{T_B}{2} = \frac{850}{2} = 425 	ext{ K}$$

Now, we can calculate the heat exchanged. At constant volume, the heat $Q$ is given by $nC_V\Delta T$. For a monoatomic gas, the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$.

$$Q_{BC} = 1 	imes \left( \frac{3}{2} 	imes 8.31 ight) 	imes (425 - 850) = -5297.6 	ext{ J}$$

The negative sign is a mathematical confirmation of what we already knew physically: heat is being lost by the gas to the surroundings.

### The Elegant Shortcut for the Final Temperature

Finally, we need to find the temperature at state $D$. While we could trudge through the pressure-temperature relations again, there is a much more elegant, "Jedi-level" shortcut.

Let's write the volume-temperature relations for the two adiabatic processes:

$$T_A V_A^{\gamma-1} = T_B V_B^{\gamma-1}$$
$$T_D V_D^{\gamma-1} = T_C V_C^{\gamma-1}$$

Now, look at the $p-V$ diagram. The processes $B ightarrow C$ and $D ightarrow A$ are isochoric. This means $V_B = V_C$ and $V_A = V_D$.

If we divide the first adiabatic equation by the second, the volume terms perfectly cancel each other out!

$$\frac{T_A}{T_D} = \frac{T_B}{T_C}$$

Rearranging for $T_D$:

$$T_D = \frac{T_A T_C}{T_B} = \frac{1000 	imes 425}{850} = 500 	ext{ K}$$

And just like that, with a simple ratio, we've found the final piece of the puzzle. This elegant symmetry is what makes thermodynamics so incredibly satisfying to study!

Visualized Solution

$Analyzing the Thermodynamic Cycle$

$Temperature at State B$

$Calculating T_{B}$

$Work Done in A \to B$

$Temperature at State C$

$Heat Lost in B \to C$

$Elegant Relation for T_{D}$

$Calculating T_{D}$

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation for Adiabatic Expansion

Calculating the Work Done

Isochoric Cooling and Heat Loss

The Elegant Shortcut for the Final Temperature

Similar Questions

In the given P-V diagram, a monoatomic gas $(γ = \frac{5}{3})$ is first compressed adiabatically from state A to state B. Then it expands isothermally from state B to state C. [Given: $(\frac{1}{3})^{0.6} ≃ 0.5, ln 2 ≃ 0.7$ ]. Which of the following statement(s) is(are) correct?

At $2 7^{\circ} C$ two moles of an ideal monoatomic gas occupy a volume $V$ . The gas expands adiabatically to a volume $2 V$ . Calculate (a) the final temperature of the gas, (b) change in its internal energy, (c) the work done by the gas during this process.

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Thermodynamic Cycle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Temperature at State B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating TB​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done in A→B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Temperature at State C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Lost in B→C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elegant Relation for TD​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating TD​