One mole of a diatomic ideal gas (gamma = 1.4) is taken through a cyclic process starting from point

Q: Why is there no heat exchange in processes $A \rightarrow B$ and $C \rightarrow D$?

By definition, an adiabatic process is one where the system is perfectly insulated, meaning no heat ($Q = 0$) can enter or leave the gas. The temperature changes purely due to the work done on or by the gas.

Q: How did we know that $V_D = V_A$?

The problem states that the process $D \rightarrow A$ is isochoric. 'Isochoric' means constant volume. Therefore, the volume at the end of the process ($V_A$) must be exactly the same as the volume at the start of the process ($V_D$).

Q: Why do we use $C_p$ for $B \rightarrow C$ and $C_V$ for $D \rightarrow A$?

The process $B \rightarrow C$ is isobaric (constant pressure), so the heat exchanged is calculated using the molar heat capacity at constant pressure, $C_p$. The process $D \rightarrow A$ is isochoric (constant volume), so we use the molar heat capacity at constant volume, $C_V$.

Q: Why is the efficiency not 100%?

According to the Second Law of Thermodynamics, no heat engine can be 100% efficient. Some heat must always be rejected to the cold reservoir (in this case, during the isochoric cooling $D \rightarrow A$) to complete the cycle and return the gas to its initial state.

Q: What if the gas was monoatomic instead of diatomic?

If the gas were monoatomic, the value of $\gamma$ would be $5/3 \approx 1.67$ instead of $1.4$. This would lead to a steeper adiabatic curve, different temperatures at the vertices, and ultimately a different efficiency for the cycle.

Question

One mole of a diatomic ideal gas ($\gamma = 1.4$) is taken through a cyclic process starting from point $A$. The process $A ightarrow B$ is an adiabatic compression. $B ightarrow C$ is isobaric expansion, $C ightarrow D$ an adiabatic expansion and $D ightarrow A$ is isochoric. The volume ratio are $V_A/V_B = 16$ and $V_C/V_B = 2$ and the temperature at $A$ is $T_A = 300	ext{ K}$. Calculate the temperature of the gas at the points $B$ and $D$ and find the efficiency of the cycle.

Sigma Percentile · Accepted Answer

### Visualizing the Thermodynamic Cycle

To truly understand a thermodynamic engine, we must first map its journey. Imagine the gas inside a cylinder undergoing a sequence of four distinct transformations. We start at state $A$. The gas is suddenly compressed without any heat exchange—an **adiabatic compression** to state $B$. The pressure spikes, and the volume shrinks.

Next, the gas is allowed to expand while the pressure is held perfectly constant. This is the **isobaric expansion** from $B$ to $C$, where the gas absorbs heat from its surroundings. Following this, the gas continues to expand, but now the insulation is back on. This **adiabatic expansion** takes it to state $D$. Finally, the volume is locked, and the gas cools down, dropping its pressure back to the initial state $A$. This is the **isochoric process**.

### Decoding the Volume Ratios

The problem provides us with crucial volume ratios: $V_A/V_B = 16$ and $V_C/V_B = 2$. Let's anchor our volumes by setting $V_B = V_0$. This immediately tells us that $V_A = 16V_0$ and $V_C = 2V_0$.

Because the final process $D ightarrow A$ is isochoric (constant volume), the volume at $D$ must be identical to the volume at $A$. Therefore, $V_D = 16V_0$. We now have a complete geometric map of the cycle's boundaries.

### Tracking the Temperatures: The Adiabatic and Isobaric Journeys

We are given the starting temperature $T_A = 300	ext{ K}$. To find the temperature at $B$, we use the adiabatic relation for temperature and volume:

$$T_A V_A^{\gamma - 1} = T_B V_B^{\gamma - 1}$$

Rearranging for $T_B$, we get:

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

Plugging in our values for a diatomic gas ($\gamma = 1.4$), we find $T_B = 300 	imes (16)^{0.4} \approx 909	ext{ K}$.

Moving from $B$ to $C$, the process is isobaric. According to Charles's Law, volume is directly proportional to temperature ($V \propto T$). Since the volume doubles from $V_0$ to $2V_0$, the temperature must also double. Thus, $T_C = 2 	imes 909 = 1818	ext{ K}$.

Finally, we traverse the adiabatic expansion from $C$ to $D$. Using the same adiabatic relation:

$$T_D = T_C \left( \frac{V_C}{V_D} ight)^{\gamma - 1} = 1818 	imes \left( \frac{2V_0}{16V_0} ight)^{0.4}$$

This simplifies to $1818 	imes (1/8)^{0.4} \approx 791.4	ext{ K}$.

### The Efficiency of the Engine

The efficiency $\eta$ of any heat engine is determined by how much of the absorbed heat is converted into useful work. Mathematically, it is expressed as:

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

In our cycle, heat is only absorbed ($Q_{	ext{in}}$) during the isobaric expansion $B ightarrow C$. Heat is only rejected ($Q_{	ext{out}}$) during the isochoric cooling $D ightarrow A$. The adiabatic processes, by definition, involve zero heat exchange.

### The Final Calculation

Let's calculate the exact heat values. For the isobaric heating of a diatomic gas, we use the molar heat capacity at constant pressure, $C_p = \frac{7}{2}R$:

$$Q_{	ext{in}} = n C_p (T_C - T_B) = (1) \left( \frac{7}{2} R ight) (1818 - 909) \approx 26438.3	ext{ J}$$

For the isochoric cooling, we use the molar heat capacity at constant volume, $C_V = \frac{5}{2}R$:

$$Q_{	ext{out}} = |n C_V (T_D - T_A)| = (1) \left( \frac{5}{2} R ight) (791.4 - 300) \approx 10208.8	ext{ J}$$

Substituting these into our efficiency formula yields:

$$\eta = 1 - \frac{10208.8}{26438.3} = 1 - 0.386 = 0.614$$

Thus, the efficiency of this thermodynamic cycle is a remarkable **$61.4\%$**.

Visualized Solution

\text{Visualizing the Cyclic Process}

\text{Decoding the Volume Ratios}

\text{Temperature at } B \text{ (Adiabatic Process)}

\text{Calculating } T_B

\text{Temperature at } C \text{ (Isobaric Process)}

\text{Temperature at } D \text{ (Adiabatic Process)}

\text{Calculating } T_D

\text{Efficiency of the Cycle}

\text{Calculating Heat Exchanged}

\text{Evaluating Heat Values}

\text{Final Efficiency}

\text{The Way Forward}

The Sigma Insight: Thermodynamic Processes

Visualizing the Thermodynamic Cycle

Decoding the Volume Ratios

Tracking the Temperatures

The Efficiency of the Engine

The Final Calculation

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