Ten moles of an ideal monoatomic gas, initially in state a at atmospheric pressure and temperature T

Q: Why is the first process considered adiabatic?

The problem states the gas is 'sudden compressed'. In a sudden process, there isn't enough time for heat to be exchanged with the surroundings, which is the defining characteristic of an adiabatic process ($Q = 0$).

Q: Why does the internal energy only depend on temperature?

For an ideal gas, there are no intermolecular forces. Therefore, the internal energy is purely kinetic and depends solely on the absolute temperature of the gas, given by $U = nC_vT$.

Q: How do we know the final process is isothermal?

The gas is expanded 'slowly' while submerged in a 'water bath'. The water bath acts as a massive heat reservoir, and the slow expansion allows the gas to constantly exchange heat to maintain thermal equilibrium with the bath, keeping the temperature constant.

Q: Why is point $f$ below point $a$ on the P-V diagram?

Both points $a$ and $f$ are at the same volume $V_0$. However, the temperature at $f$ ($284\text{ K}$) is lower than at $a$ ($300\text{ K}$). According to the ideal gas law ($P = nRT/V$), at constant volume, a lower temperature results in a lower pressure.

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

The problem specifies that the gas is 'monoatomic'. For monoatomic gases (like Helium or Argon), the degrees of freedom are 3 (translational only), which gives $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$. The ratio $\gamma = C_p/C_v$ is therefore $5/3$.

Question

Ten moles of an ideal monoatomic gas, initially in state $a$ at atmospheric pressure and temperature $T_a = 27^\circ	ext{C}$, is enclosed in a metal cylinder of volume $V_0$ fitted with a frictionless piston. The gas is suddenly compressed to state $b$ with volume $V_0/3$. Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature $11^\circ	ext{C}$ until the gas reaches the temperature of the water bath, which is denoted as state $c$. Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state $f$. If $R$ is universal gas constant, then the correct option(s) is/are: [Given: $9^{1/3} = 2.08$]

Options:

(A) The schematic P-V diagram of the processes described above is:

(B) The change in internal energy in going from state $a$ to $b$ is $4860R$.

(C) The net change in the internal energy in the whole process is $-240R$.

(D) The pressure and temperature of the state $b$ are $2.08$ times the atmospheric pressure and $624	ext{ K}$, respectively.

Sigma Percentile · Accepted Answer

### The Setup: A Thermodynamic Journey

Imagine you are an engineer tasked with controlling the state of ten moles of an ideal monoatomic gas. This gas is trapped inside a metal cylinder with a frictionless piston, initially resting comfortably at state $a$ with atmospheric pressure and a temperature of $27^\circ	ext{C}$ (which is $300	ext{ K}$).

Our journey involves three distinct phases, each with its own unique physical constraints. The problem tests our ability to translate physical descriptions into mathematical thermodynamic processes.

### Phase 1: The Sudden Squeeze

The first action is a **sudden compression** to a third of its original volume, bringing us to state $b$. In physics, the word "sudden" is a massive clue. It implies that the process happens so fast that there is no time for heat exchange with the surroundings. This makes it an **adiabatic process**.

For a monoatomic gas, the adiabatic index $\gamma$ is $\frac{5}{3}$. We can find the new temperature $T_b$ using the adiabatic relation between temperature and volume:

$$T_a V_a^{\gamma-1} = T_b V_b^{\gamma-1}$$

Substituting our known values:

$$300 \cdot V_0^{\frac{5}{3}-1} = T_b \cdot \left(\frac{V_0}{3}ight)^{\frac{5}{3}-1}$$

$$300 \cdot V_0^{\frac{2}{3}} = T_b \cdot \frac{V_0^{\frac{2}{3}}}{3^{\frac{2}{3}}}$$

The $V_0$ terms cancel out beautifully, leaving us with:

$$T_b = 300 \cdot 3^{\frac{2}{3}} = 300 \cdot (3^2)^{\frac{1}{3}} = 300 \cdot 9^{\frac{1}{3}}$$

Given that $9^{\frac{1}{3}} = 2.08$, we find:

$$T_b = 300 \cdot 2.08 = 624	ext{ K}$$

Now, let's check the pressure at state $b$. Using the relation $P_a V_a^\gamma = P_b V_b^\gamma$:

$$P_0 V_0^{\frac{5}{3}} = P_b \left(\frac{V_0}{3}ight)^{\frac{5}{3}}$$

$$P_b = P_0 \cdot 3^{\frac{5}{3}} = P_0 \cdot 3 \cdot 3^{\frac{2}{3}} = 3 \cdot 2.08 P_0 = 6.24 P_0$$

Option (D) claims the pressure is $2.08$ times the atmospheric pressure, which is a trap! It's actually $6.24$ times. So, option (D) is incorrect.

What about the energy? The change in internal energy during this adiabatic compression is given by:

$$\Delta U_{ab} = n C_v \Delta T$$

For our monoatomic gas, $C_v = \frac{3}{2}R$. Plugging in the numbers:

$$\Delta U_{ab} = 10 \cdot \frac{3}{2}R \cdot (624 - 300) = 15R \cdot 324 = 4860R$$

This perfectly matches option (B)!

### Phase 2 & 3: Cooling and Expanding

Next, the piston is kept stationary. A locked piston means the volume cannot change, making this an **isochoric process**. The cylinder is submerged in a water bath at $11^\circ	ext{C}$ ($284	ext{ K}$), so the gas cools down to this temperature at state $c$.

Finally, the piston is brought slowly back to its initial position while still in the water bath. The word "slowly" combined with the water bath ensures that the gas remains in thermal equilibrium with the bath. This is an **isothermal process**. Therefore, the final temperature $T_f$ at state $f$ is also $284	ext{ K}$.

### The Grand Finale: Net Energy and the P-V Map

To find the net change in internal energy for the entire cycle, we must remember a fundamental principle: **internal energy is a state function**. It doesn't care about the path taken; it only depends on the final and initial states!

$$\Delta U_{net} = U_f - U_a = n C_v (T_f - T_a)$$

$$\Delta U_{net} = 10 \cdot \frac{3}{2}R \cdot (284 - 300) = 15R \cdot (-16) = -240R$$

This confirms that option (C) is absolutely correct.

Lastly, let's evaluate the P-V diagram in option (A). 
1. $a 	o b$: A steep curve upwards and leftwards, characteristic of an adiabatic compression.
2. $b 	o c$: A straight vertical drop, representing the isochoric cooling.
3. $c 	o f$: A gentler curve downwards and rightwards for the isothermal expansion.

Crucially, because the final temperature $T_f$ ($284	ext{ K}$) is lower than the initial temperature $T_a$ ($300	ext{ K}$), the final pressure $P_f$ must be lower than the initial pressure $P_a$. The diagram correctly shows point $f$ lying below point $a$. Thus, the schematic is accurate, and option (A) is correct!

Select Answer:

Visualized Solution

\text{Thermodynamic Cycle Overview}

\text{Process } a \to b: \text{Adiabatic Compression}

\text{Calculating } T_b

\text{Calculating } P_b \text{ (Checking Option D)}

\text{Change in Internal Energy } (\Delta U_{ab})

\text{Process } b \to c \text{ and } c \to f

\text{Net Change in Internal Energy } (\Delta U_{net})

\text{Verifying the P-V Diagram}

The Sigma Insight: Thermodynamic Processes

The Setup

Phase 1

Phase 2 & 3

The Grand Finale

Similar Questions

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List-I

List-II

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