Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. T

Q: Why is the process in cylinder A at constant pressure?

Because the piston is free to move, it will adjust its position to maintain mechanical equilibrium with the constant external atmospheric pressure (and its own weight). Thus, the gas pressure remains constant during heating.

Q: Why does cylinder B have a higher temperature rise?

In cylinder A, some of the supplied heat is used to do work (expanding against the piston). In cylinder B, the volume is fixed, so no work is done. All the supplied heat goes entirely into increasing the internal energy, resulting in a higher temperature rise.

Q: What is the value of $\gamma$ for a diatomic gas?

For an ideal diatomic gas at normal temperatures, the degrees of freedom $f = 5$ (3 translational + 2 rotational). The adiabatic index $\gamma = 1 + 2/f = 1 + 2/5 = 7/5 = 1.4$.

Q: Does the initial temperature of 300 K matter in this calculation?

No, the initial temperature is extra information. The relationship between the temperature changes ($\Delta T$) depends only on the specific heats and the fact that the heat supplied is equal.

Q: What if the gas was monoatomic?

For a monoatomic gas, $\gamma = 5/3$. The temperature rise in cylinder B would be $\Delta T_B = (5/3) \times 30 = 50\text{ K}$.

Question

Two cylinders $A$ and $B$ fitted with pistons contain equal amounts of an ideal diatomic gas at $300	ext{ K}$. The piston of $A$ is free to move, while that of $B$ is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in $A$ is $30	ext{ K}$, then the rise in temperature of the gas in $B$ is

Options:

(A) $30	ext{ K}$

(B) $18	ext{ K}$

(C) $50	ext{ K}$

(D) $42	ext{ K}$

Sigma Percentile · Accepted Answer

The problem presents us with a fascinating scenario involving two identical cylinders, $A$ and $B$, both filled with the same amount of an ideal diatomic gas at an initial temperature of $300	ext{ K}$. The key difference lies in their pistons: cylinder $A$'s piston is free to move, while cylinder $B$'s piston is rigidly fixed.

When we supply the exact same amount of heat to both cylinders, they respond differently based on their mechanical constraints. Let's break down the physics behind this.

### Analyzing the Setup

For cylinder $A$, the piston is free to move. This means that as the gas is heated, it will expand against the constant external atmospheric pressure (and the weight of the piston). Because the pressure remains constant, this is an **isobaric process**. The heat supplied to the gas in cylinder $A$ can be expressed using the molar specific heat at constant pressure, $C_p$:

$$ \Delta Q_A = n C_p \Delta T_A $$

For cylinder $B$, the piston is held fixed. The gas cannot expand, meaning its volume remains constant. This is an **isochoric process**. The heat supplied to the gas in cylinder $B$ is expressed using the molar specific heat at constant volume, $C_V$:

$$ \Delta Q_B = n C_V \Delta T_B $$

### The Master Equation

The problem explicitly states that the same amount of heat is given to the gas in each cylinder. Therefore, we can equate the two heat expressions:

$$ \Delta Q_A = \Delta Q_B $$

Substituting our formulas into this equality, we get:

$$ n C_p \Delta T_A = n C_V \Delta T_B $$

Notice how the number of moles, $n$, is the same on both sides. We can safely cancel it out. We are looking for the rise in temperature of the gas in cylinder $B$, which is $\Delta T_B$. Rearranging the equation to solve for $\Delta T_B$, we find:

$$ \Delta T_B = \left(\frac{C_p}{C_V}ight) \Delta T_A $$

The ratio of the molar specific heats, $\frac{C_p}{C_V}$, is a very important thermodynamic property known as the **adiabatic index**, denoted by the Greek letter $\gamma$. So, our equation simplifies beautifully to:

$$ \Delta T_B = \gamma \Delta T_A $$

### Final Calculation

Now, we need to determine the value of $\gamma$. The problem tells us that the gas is **diatomic**. For an ideal diatomic gas at normal temperatures, the adiabatic index is a standard known value:

$$ \gamma = 1.4 $$

We are also given that the rise in temperature of the gas in cylinder $A$ is $\Delta T_A = 30	ext{ K}$. Substituting these values into our master equation:

$$ \Delta T_B = 1.4 	imes 30	ext{ K} $$

$$ \Delta T_B = 42	ext{ K} $$

**The rise in temperature of the gas in cylinder $B$ is $42	ext{ K}$.**

This result makes perfect physical sense. In cylinder $A$, some of the supplied heat energy is "wasted" doing mechanical work to push the piston up. In cylinder $B$, no work is done, so 100% of the supplied heat goes directly into increasing the internal kinetic energy of the gas molecules, resulting in a significantly higher temperature rise!

Select Answer:

Visualized Solution

$Analyzing the Setup$

$Heat Supplied Equations$

$Equating the Heat$

$Rearranging for Δ T_{B}$

$Substituting Values$

$Final Calculation$

$What if it was Monoatomic?$

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

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; } Analyzing the Setup

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if it was Monoatomic?