1 mole of rigid diatomic gas performs a work of Q/5 when heat Q is supplied to it. The molar heat ca

Q: Why do we use $C_V$ for internal energy even if the volume is changing?

The internal energy of an ideal gas depends ONLY on its temperature. The formula $\Delta U = n C_V \Delta T$ is universally true for any process involving an ideal gas, regardless of whether the volume changes or not.

Q: What does 'rigid' diatomic gas mean?

'Rigid' means the bond length between the two atoms is fixed, so vibrational degrees of freedom are not excited. It only has 3 translational and 2 rotational degrees of freedom, making $f = 5$.

Q: Is this a polytropic process?

Yes! Any process where the molar heat capacity $C$ is constant is a polytropic process. Here, $C = \frac{25R}{8}$, which is a constant, so it follows the equation $PV^n = \text{constant}$.

Q: How did we know $n=1$?

The question explicitly states '1 mole of rigid diatomic gas', so we directly substitute $n=1$ into our equations.

Q: Can we solve this by finding the polytropic index $n$?

Absolutely! You can use $W = \frac{nR\Delta T}{1-n}$ and $Q = nC\Delta T$ to find the polytropic index, and then use $C = C_V + \frac{R}{1-n}$. Both methods will yield the exact same result.

Question

1 mole of rigid diatomic gas performs a work of $\frac{Q}{5}$ when heat $Q$ is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{xR}{8}$, The value of $x$ is ........... .
$[R = 	ext{universal gas constant}]$

Sigma Percentile · Accepted Answer

### Setting the Stage: The Thermodynamic Process

Imagine you are observing a perfectly sealed cylinder fitted with a movable piston. Inside this cylinder, we have exactly **one mole of a rigid diatomic gas**.

The term "rigid" is crucial here—it tells us that the bond between the two atoms is fixed, meaning the molecules can translate and rotate, but they cannot vibrate. This locks our degree of freedom, $f$, at exactly 5.

Now, we ignite a heat source. We supply a total heat energy of $Q$ to the system.

As the gas absorbs this heat, it gets energized, expands, and pushes the piston upwards. The problem tells us that the work done by the gas during this expansion is exactly $W = \frac{Q}{5}$.

### The First Law: Balancing the Energy Checkbook

To understand what happens to the rest of the energy, we must invoke the ultimate accountant of the universe: **The First Law of Thermodynamics**.

$$ \Delta Q = \Delta U + \Delta W $$

This law simply states that the heat supplied to a system is used to do external work and to increase the system's internal energy. Let's plug in our known values:

$$ Q = \Delta U + \frac{Q}{5} $$

By rearranging this equation, we can easily find the change in the internal energy of the gas:

$$ \Delta U = Q - \frac{Q}{5} = \frac{4Q}{5} $$

This tells us that 80% of the heat supplied goes into raising the temperature of the gas, while the remaining 20% is used to do work.

### Diving into Internal Energy

We know that for any ideal gas, the change in internal energy is strictly a function of temperature. The universal formula is:

$$ \Delta U = n C_V \Delta T $$

Since we are dealing with a rigid diatomic gas, its molar heat capacity at constant volume, $C_V$, is given by $\frac{f}{2}R$. With $f = 5$, we have $C_V = \frac{5}{2}R$.

We also know that we have exactly 1 mole of gas ($n = 1$). Let's substitute these into our internal energy equation:

$$ \Delta U = (1) \left(\frac{5}{2}R\right) \Delta T $$

Now, we have two distinct expressions for $\Delta U$. Let's equate them to bridge the gap between the heat supplied and the temperature change:

$$ \frac{4Q}{5} = \frac{5}{2}R \Delta T $$

Our goal is to express the heat $Q$ in terms of the temperature change $\Delta T$. By cross-multiplying, we get:

$$ Q = \frac{25}{8}R \Delta T $$

### The Grand Finale: Finding the Molar Heat Capacity

We are now in the final stretch. The question asks for the molar heat capacity, $C$, of the gas during this specific transformation.

By definition, molar heat capacity is the amount of heat required to raise the temperature of one mole of a substance by one degree:

$$ C = \frac{Q}{n \Delta T} $$

Let's substitute our expression for $Q$ and set $n = 1$:

$$ C = \frac{\frac{25}{8}R \Delta T}{(1) \Delta T} $$

Notice how beautifully the $\Delta T$ terms cancel out, leaving us with a constant value:

$$ C = \frac{25}{8}R $$

The problem states that the molar heat capacity is $\frac{xR}{8}$. By directly comparing our result with the given expression, the answer becomes crystal clear.

**The value of $x$ is 25.**

This process, where the molar heat capacity remains constant, is a classic example of a **polytropic process**. You could also solve this by finding the polytropic index $n$, but the energy balance method we used is far more intuitive and elegant!

Animated Solution for Physics - Thermodynamics: 1 mole of rigid diatomic gas performs a work of $\frac{Q}{5}$ when heat $Q$ is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{x R}{8}$ , The value of $x$ is ........... . $[R = universal gas constant]$

Enter Numerical Value:

Visualized Solution

$Visualizing the System$

$Energy Exchange$

$First Law of Thermodynamics$

$Change in Internal Energy$

$Internal Energy of Diatomic Gas$

$Equating Internal Energy$

$Expressing Heat Q$

$Molar Heat Capacity$

$Final Answer$

$The Way Forward$

The Sigma Insight: First Law of Thermodynamics

Setting the Stage

The First Law

Diving into Internal Energy

The Grand Finale

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Enter Numerical Value:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Law of Thermodynamics

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Change in Internal Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Internal Energy of Diatomic Gas

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

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

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

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

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