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

Q: Why do we assume the process is adiabatic just because it's sudden?

Heat transfer requires time. When a piston is released suddenly, the gas expands almost instantaneously. The time scale of the expansion is much smaller than the time scale required for significant heat conduction to occur across the cylinder walls. Thus, $Q = 0$.

Q: What happened to the cross-sectional area $A$ in the equation?

The area $A$ is constant. When we write $T(A \cdot l)^{\gamma-1} = \text{constant}$, we can rewrite it as $T \cdot l^{\gamma-1} \cdot A^{\gamma-1} = \text{constant}$. Dividing both sides by the constant $A^{\gamma-1}$ just gives us a new constant. So, $T \cdot l^{\gamma-1}$ is also constant.

Q: How do we know $\gamma = 5/3$ for this gas?

The problem states it's a monoatomic ideal gas. Monoatomic gases have 3 translational degrees of freedom ($f=3$). The formula for $\gamma$ is $1 + 2/f$. Plugging in $f=3$ gives $\gamma = 1 + 2/3 = 5/3$.

Q: Would the answer change if the gas was diatomic, like Oxygen?

Absolutely! For a diatomic gas at room temperature, the degrees of freedom are 5 (3 translational + 2 rotational). This makes $\gamma = 1 + 2/5 = 7/5$. The exponent in our final answer would then be $7/5 - 1 = 2/5$.

Q: Why does the temperature decrease during this expansion?

In an adiabatic expansion, the gas does positive work on the piston to push it up. Since no heat is added to the system ($Q=0$), the First Law of Thermodynamics ($\Delta U = Q - W$) tells us that the internal energy must decrease ($\Delta U = -W$). A decrease in internal energy directly corresponds to a drop in temperature.

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 the expansion respectively, then the value of $\frac{T_1}{T_2}$ will be

Options:

(A) $\left(\frac{l_1}{l_2}\right)^{\frac{2}{3}}$

(B) $\left(\frac{l_2}{l_1}\right)^{\frac{2}{3}}$

(C) $\frac{l_2}{l_1}$

(D) $\frac{l_1}{l_2}$

Sigma Percentile · Accepted Answer

### The Magic of Sudden Expansion

Imagine you are holding a bicycle pump, and you block the nozzle with your thumb. If you press the piston down really fast, the air inside gets hot. Conversely, if you let the compressed air expand suddenly, it cools down. This rapid change is the heart of our problem.

When the problem states that the piston is released **suddenly**, it is giving us a massive hint. A sudden process in thermodynamics means there is absolutely no time for heat to flow into or out of the gas. The system is perfectly insulated by its own speed! This makes the process **adiabatic**.

### The Master Equation

For an adiabatic process involving an ideal gas, the relationship between temperature $T$ and volume $V$ is governed by a beautiful equation:
$$TV^{\gamma-1} = 	ext{constant}$$

Here, $\gamma$ (gamma) is the ratio of specific heats ($C_p/C_v$), and its value depends entirely on the atomicity of the gas.

But wait, our question doesn't talk about volume; it talks about the length of the gas column, $l$. How do we bridge this gap?

### Geometry Meets Thermodynamics

Think about the shape of the cylinder. The volume $V$ of a cylinder is simply its cross-sectional area $A$ multiplied by its length $l$. 
$$V = A \cdot l$$

Let's substitute this geometric reality into our thermodynamic equation:
$$T(A \cdot l)^{\gamma-1} = 	ext{constant}$$

Now, here is the elegant part. The cross-sectional area $A$ of the cylinder is a constant. It doesn't change as the piston moves up or down. Therefore, $A^{\gamma-1}$ is also just another constant. We can divide the right side by this constant to get a brand new, simplified constant!
$$T \cdot l^{\gamma-1} = \frac{	ext{constant}}{A^{\gamma-1}} = 	ext{constant}'$$

### The Final Calculation

Now we have a direct relationship between the temperature and the length of the column:
$$T_1 l_1^{\gamma-1} = T_2 l_2^{\gamma-1}$$

We need to find the ratio of the initial temperature to the final temperature, $\frac{T_1}{T_2}$. Rearranging our equation gives:
$$\frac{T_1}{T_2} = \left(\frac{l_2}{l_1}ight)^{\gamma-1}$$

The final piece of the puzzle is the gas itself. The problem explicitly states it is a **monoatomic ideal gas** (like Helium or Argon). For a monoatomic gas, the degrees of freedom are 3, which gives us:
$$\gamma = \frac{5}{3}$$

Let's plug this into our exponent:
$$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$$

Substituting this back into our ratio equation, we arrive at the final, elegant result:
$$\frac{T_1}{T_2} = \left(\frac{l_2}{l_1}ight)^{\frac{2}{3}}$$

And there we have it! By understanding the physical meaning of "sudden" and connecting simple geometry to thermodynamic laws, we've cracked the problem.

Select Answer:

Visualized Solution

Visualizing the States

Identifying the Process

Volume Substitution

Simplifying the Equation

Applying to States

Atomicity of Gas

Final Result

The Way Forward

The Sigma Insight: Thermodynamic Processes

The Magic of Sudden Expansion

The Master Equation

Geometry Meets Thermodynamics

The Final Calculation

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