A diatomic gas having Cp = 7/2 R and CV = 5/2 R, is heated at constant pressure. The ratio dU : dQ :

Q: Why do we use $dU = nC_V dT$ even though the process is at constant pressure?

This is a very common point of confusion! For an ideal gas, the internal energy $U$ depends *only* on its temperature, not on its volume or pressure. Therefore, the formula $dU = nC_V dT$ is universally true for any process involving an ideal gas, whether it's isochoric, isobaric, or anything else.

Q: Can I calculate the work done using $dW = P dV$ instead of $nR dT$?

Absolutely! In fact, $dW = P dV$ is the fundamental definition of boundary work. However, since we are looking for a ratio involving $dT$, we use the ideal gas law ($PV = nRT$). Differentiating at constant pressure gives $P dV = nR dT$. Substituting this makes it easy to cancel out $dT$ in the ratio.

Q: What would the ratio be if the gas was monatomic instead of diatomic?

For a monatomic gas, the specific heats are different because the molecules only have translational kinetic energy. $C_V = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$. Following the exact same logic, the ratio $dU : dQ : dW$ would be $\frac{3}{2}R : \frac{5}{2}R : R$, which simplifies to $3 : 5 : 2$.

Q: Is it a coincidence that the numbers in the ratio $5:7:2$ relate to each other?

Not at all! It's a direct consequence of Mayer's relation: $C_p - C_V = R$. If you look at the ratio $C_V : C_p : R$, it's essentially $C_V : (C_V + R) : R$. In our case, $5/2 : 7/2 : 2/2$, you can see that $5 + 2 = 7$, which perfectly mirrors the First Law: $dU + dW = dQ$.

Q: What if the process was isochoric (constant volume) instead of isobaric?

In an isochoric process, the volume doesn't change, so the gas does zero work ($dW = 0$). All the heat supplied goes directly into increasing the internal energy. Therefore, $dQ = dU$, and the ratio $dU : dQ : dW$ would simply be $1 : 1 : 0$.

Question

A diatomic gas having $C_p = \frac{7}{2}R$ and $C_V = \frac{5}{2}R$, is heated at constant pressure. The ratio $dU : dQ : dW$ is

Options:

(A) 5 : 7 : 3

(B) 5 : 7 : 2

(C) 3 : 7 : 2

(D) 3 : 5 : 2

Sigma Percentile · Accepted Answer

### The Thermodynamics of Heating a Gas

Imagine you are conducting an experiment in a lab. You have a cylinder filled with a diatomic gas, like Oxygen or Nitrogen, and it's sealed with a frictionless, movable piston. You start heating the cylinder. Because the piston is free to move, the gas expands as it gets hotter, maintaining a constant pressure against the atmosphere. This is the classic **isobaric process**.

But what exactly happens to the heat energy you are pumping into the system? According to the **First Law of Thermodynamics**, energy cannot be destroyed; it must be accounted for. The heat supplied ($dQ$) splits into two distinct jobs:
1.  **Increasing the Internal Energy ($dU$):** This makes the gas molecules move and rotate faster, raising the temperature.
2.  **Doing Mechanical Work ($dW$):** This is the energy spent by the gas to push the piston outward against the external pressure.

### The Master Equations

To find the ratio of how this energy is partitioned, we need the mathematical expressions for each component. For an ideal gas undergoing a small temperature change $dT$:

*   The change in internal energy is always given by $dU = n C_V dT$, regardless of the process. Here, $C_V$ is the molar specific heat at constant volume.
*   The total heat supplied at constant pressure is $dQ = n C_p dT$, where $C_p$ is the molar specific heat at constant pressure.
*   The work done by the gas is $dW = P dV$. Using the ideal gas equation ($PV = nRT$), at constant pressure, $P dV = nR dT$. So, $dW = nR dT$.

### Calculating the Energy Partition Ratio

We are asked to find the ratio $dU : dQ : dW$. Let's substitute our expressions into this ratio:

$$dU : dQ : dW = n C_V dT : n C_p dT : n R dT$$

Notice something beautiful? The number of moles ($n$) and the temperature change ($dT$) are common factors in all three terms. They cancel out perfectly, leaving us with a profound and simple relationship:

$$dU : dQ : dW = C_V : C_p : R$$

This tells us that for *any* ideal gas heated at constant pressure, the way energy is distributed is entirely dictated by its specific heats!

### The Final Calculation

For our specific problem, we are dealing with a diatomic gas. The problem provides the specific heats:
*   $C_V = \frac{5}{2}R$
*   $C_p = \frac{7}{2}R$

Substituting these into our simplified ratio:

$$dU : dQ : dW = \frac{5}{2}R : \frac{7}{2}R : R$$

To clean up this ratio and remove the fractions, we can divide every term by $R$ and multiply by $2$.

$$dU : dQ : dW = 5 : 7 : 2$$

This means for every $7$ Joules of heat you supply, $5$ Joules go into heating up the gas (internal energy), and $2$ Joules are used to push the piston (work done). It's a perfect, elegant balance dictated by the fundamental nature of the diatomic molecules!

Animated Solution for Physics - Thermodynamics: A diatomic gas having $C_{p} = \frac{7}{2} R$ and $C_{V} = \frac{5}{2} R$ , is heated at constant pressure. The ratio $d U : d Q : d W$ is

Select Answer:

Visualized Solution

Isobaric Heating

The First Law Components

Substituting Specific Heats

Simplifying the Ratio

Final Ratio

Conceptual Takeaway

The Sigma Insight: First Law of Thermodynamics

The Thermodynamics of Heating a Gas

The Master Equations

Calculating the Energy Partition Ratio

The Final Calculation

Similar Questions

$n$ 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

A reversible cyclic process for an ideal gas is shown below. Here, P , V and T are pressure , volume and temperature , respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

5 moles of an ideal gas at $100 K$ are allowed to undergo reversible compression till its temperature becomes $200 K$ . If $C_{V} = 28 JK^{- 1} mol^{- 1}$ , calculate $Δ U$ and $Δ (p V)$ for this process. ( $R = 8.0 JK^{- 1} mol^{- 1}$ )

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 ........... . $[R = universal gas constant]$

Following figure shows two processes A and B for a gas. If $Δ Q_{A}$ and $Δ Q_{B}$ are the amount of heat absorbed by the system in two cases, and $Δ U_{A}$ and $Δ U_{B}$ are changes in internal energies respectively, then

Five moles of an ideal gas at $1 bar$ and $298 K$ is expanded into vacuum to double the volume. The work done is

For an ideal gas

In a given process of an ideal gas, $d W = 0$ and $d Q < 0$ . Then for the gas

A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?

Animated Solution for Physics - Thermodynamics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A diatomic gas having Cp​=27​R and CV​=25​R, is heated at constant pressure. The ratio dU:dQ:dW is