A monoatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictio

Q: Why is the process considered adiabatic?

The word "sudden" implies that the expansion happens so fast that there is no time for heat to flow into or out of the gas ($Q = 0$).

Q: Isn't a sudden expansion irreversible? Why use $TV^{\gamma-1} = \text{constant}$?

Strictly speaking, a sudden expansion against a constant external pressure is irreversible. However, in standard JEE problems, if external pressure isn't provided, "sudden" is merely a cue for "adiabatic", and we use the reversible equations as an expected approximation.

Q: What is $\gamma$ and why is it $5/3$?

$\gamma$ is the ratio of specific heats ($C_p/C_v$). For a monoatomic gas, it has 3 translational degrees of freedom, giving $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$, so $\gamma = 5/3$.

Q: What happens to the cross-sectional area $A$?

The volume of the cylinder is $V = A \times L$. When we take the ratio $V_2/V_1$, the constant cross-sectional area $A$ cancels out, leaving just the ratio of the lengths $L_2/L_1$.

Q: Does the temperature increase or decrease?

During an adiabatic expansion, the gas does work at the expense of its internal energy. Therefore, the internal energy decreases, which means the temperature drops ($T_2 < T_1$).

Question

A monoatomic ideal gas, initially at temperature $T_1$, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature $T_2$ by releasing the piston suddenly. If $L_1$ and $L_2$ are the lengths of the gas column before and after expansion respectively, then $T_1/T_2$ is given by

Options:

(A) $\left(\frac{L_1}{L_2}\right)^{2/3}$

(B) $\frac{L_1}{L_2}$

(C) $\frac{L_2}{L_1}$

(D) $\left(\frac{L_2}{L_1}\right)^{2/3}$

Sigma Percentile · Accepted Answer

The problem presents us with a classic thermodynamic scenario: a monoatomic ideal gas enclosed in a cylinder, expanding suddenly.

The word **sudden** is our biggest clue here. In thermodynamics, a sudden process implies that it happens so quickly that there is no time for heat exchange with the surroundings. Therefore, the process is **adiabatic** ($Q = 0$).

### The Master Equation

For an adiabatic process involving an ideal gas, the relationship between temperature $T$ and volume $V$ is governed by the equation:

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

where $\gamma$ is the ratio of specific heats ($C_p/C_v$).

Applying this principle to our initial and final states, we can set up the following equality:

$$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$$

Our goal is to find the ratio of the initial temperature to the final temperature, $T_1/T_2$. Let's rearrange our equation to isolate this ratio:

$$\frac{T_1}{T_2} = \left(\frac{V_2}{V_1}ight)^{\gamma - 1}$$

### Connecting Volume to Length

The gas is enclosed in a cylinder. The volume of a cylinder is simply its cross-sectional area $A$ multiplied by its length $L$.

So, we can express the initial and final volumes as:

$$V_1 = A L_1$$

$$V_2 = A L_2$$

Substituting these into our temperature ratio equation, we get:

$$\frac{T_1}{T_2} = \left(\frac{A L_2}{A L_1}ight)^{\gamma - 1}$$

Notice how beautifully the cross-sectional area $A$ cancels out! This leaves us with a ratio that depends entirely on the lengths:

$$\frac{T_1}{T_2} = \left(\frac{L_2}{L_1}ight)^{\gamma - 1}$$

### The Monoatomic Factor

The final piece of the puzzle lies in the nature of the gas. We are told it is a **monoatomic** ideal gas.

For a monoatomic gas, the degrees of freedom are purely translational ($f = 3$). This gives us a specific heat ratio of:

$$\gamma = \frac{5}{3}$$

Now, we just need to calculate the exponent $\gamma - 1$:

$$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$$

### Final Calculation

Substituting this exponent back into our equation, we arrive at the final, elegant result:

$$\frac{T_1}{T_2} = \left(\frac{L_2}{L_1}ight)^{2/3}$$

This matches option (d). The beauty of this problem lies in recognizing the adiabatic nature of the sudden expansion and seamlessly connecting the macroscopic geometry of the cylinder to the thermodynamic state variables.

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Visualized Solution

$Visualizing the Expansion$

$Adiabatic Process Equation$

$Applying to Initial and Final States$

$Rearranging for Temperature Ratio$

$Substituting Volume in terms of Length$

$Monoatomic Gas Constant$

$Final Answer$

The Sigma Insight: Thermodynamic Processes

The Master Equation

Connecting Volume to Length

The Monoatomic Factor

Final Calculation

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adiabatic Process Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying to Initial and Final States

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging for Temperature Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Volume in terms of Length

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Monoatomic Gas Constant

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