n moles of an ideal gas with constant volume heat capacity C_V undergo an isobaric expansion by cert

Q: Why is $\Delta U = C_V \Delta T$ even for an isobaric process?

The internal energy of an ideal gas depends **only** on its temperature. Regardless of the process (isobaric, isochoric, etc.), the change in internal energy is always given by $\Delta U = C_V \Delta T$.

Q: What is the difference between $C_V$ and $c_V$?

$C_V$ usually represents the **total** heat capacity of the system, while $c_V$ represents the **molar** heat capacity. The question explicitly defines $C_V$ as the heat capacity of the $n$ moles, so we don't multiply it by $n$ again.

Q: Can we use $C_p$ to solve this?

Absolutely! The heat supplied in an isobaric process is $\Delta Q = C_p \Delta T$. Since $C_p - C_V = nR$ (for total heat capacities), we get $C_p = C_V + nR$. Substituting this gives the exact same result.

Q: Why is the work done equal to $nR\Delta T$?

For an isobaric (constant pressure) process, work done is $W = p\Delta V$. Using the ideal gas equation $pV = nRT$, at constant pressure, a change in volume $\Delta V$ corresponds to a change in temperature $\Delta T$, so $p\Delta V = nR\Delta T$.

Q: What if the gas was specified as diatomic?

The general formula $\frac{nR}{C_V + nR}$ would still hold. However, you could substitute $C_V = \frac{5}{2}nR$ to get a numerical ratio of $\frac{nR}{\frac{5}{2}nR + nR} = \frac{2}{7}$.

Question

$n$ moles of an ideal gas with constant volume heat capacity $C_V$ undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is

Options:

(A) $\frac{4nR}{C_V + nR}$

(B) $\frac{4nR}{C_V - nR}$

(C) $\frac{nR}{C_V - nR}$

(D) $\frac{nR}{C_V + nR}$

Sigma Percentile · Accepted Answer

The beauty of thermodynamics lies in its universal accounting system: the First Law. In this problem, we are asked to find the ratio of the work done by a gas to the heat supplied to it during an isobaric expansion.

Let's break down the physical reality of this process and see how the math naturally unfolds.

### Analyzing the Setup

Imagine a gas trapped in a cylinder with a movable piston. We are told the gas undergoes an **isobaric expansion**. This means we are slowly adding heat to the gas, causing it to expand and push the piston outward, all while maintaining a constant pressure.

There is a subtle but critical trap in the problem statement. The question defines $C_V$ as the "constant volume heat capacity" for the $n$ moles of gas, not the *molar* heat capacity. This means $C_V$ already accounts for the total amount of gas.

### The Master Equation

To connect heat, work, and internal energy, we rely on the **First Law of Thermodynamics**:

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

This equation is simply the conservation of energy. The heat $\Delta Q$ we supply to the gas goes into two places: increasing the internal energy $\Delta U$ (making the gas molecules move faster) and doing mechanical work $W$ (pushing the piston).

### Calculating Work and Internal Energy

Let's express both the work done and the change in internal energy in terms of the temperature change $\Delta T$.

For an isobaric process, the work done is the constant pressure multiplied by the change in volume:

$$ W = p\Delta V $$

Using the ideal gas law ($pV = nRT$), we can rewrite this in terms of temperature:

$$ W = nR\Delta T $$

Next, we look at the internal energy. The internal energy of an ideal gas depends *only* on its temperature. Regardless of the process, the change in internal energy is always the total constant volume heat capacity multiplied by the temperature change:

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

### Final Calculation

Now, we substitute these expressions back into our First Law equation to find the total heat supplied:

$$ \Delta Q = C_V\Delta T + nR\Delta T $$

Factoring out the common $\Delta T$ term, we get:

$$ \Delta Q = (C_V + nR)\Delta T $$

We are finally ready to find the requested ratio of work done to heat supplied:

$$ 	ext{Ratio} = \frac{W}{\Delta Q} = \frac{nR\Delta T}{(C_V + nR)\Delta T} $$

The $\Delta T$ terms perfectly cancel out from the numerator and the denominator, leaving us with a beautifully simple, temperature-independent ratio:

$$ 	ext{Ratio} = \frac{nR}{C_V + nR} $$

This elegant result perfectly matches option (d). It tells us exactly what fraction of our inputted heat is converted into useful mechanical work!

Animated Solution for Physics - Thermodynamics: $n$ moles of an ideal gas with constant volume heat capacity $C_{V}$ undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is

Select Answer:

Visualized Solution

Isobaric Expansion

First Law of Thermodynamics

Work and Internal Energy

Heat Supplied

Ratio of Work to Heat

Final Result

Food for Thought

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Calculating Work and Internal Energy

Final Calculation

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