Question

For diatomic gas, find the ratio $Q : \Delta U : W$ for isobaric process :-

• A. 2 : 5 : 7
• B. 2 : 3 : 5
• C. 2 : 7 : 3
• D. 7 : 5 : 2

Answer 1

The Correct Option is D

To find the ratio \( Q : \Delta U : W \) for an isobaric process involving a diatomic gas, we start by understanding the concepts involved:

1. Isobaric Process: This is a thermodynamic process in which the pressure remains constant.
2. Diatomic Gas: For a diatomic ideal gas, the degree of freedom is 5, and the specific heat capacities are \( C_p = \frac{7}{2} R \) and \( C_v = \frac{5}{2} R \).

In an isobaric process, the following relation holds true according to the first law of thermodynamics:

\(Q = \Delta U + W\)

Where,

• \( Q \) is the heat added to the system.
• \( \Delta U \) is the internal energy change.
• \( W \) is the work done by the system.

For a diatomic gas undergoing an isobaric process:

• Heat Added (Q): \( Q = n \cdot C_p \cdot \Delta T \)
• Change in Internal Energy (\(\Delta U\)): \( \Delta U = n \cdot C_v \cdot \Delta T \)
• Work Done (W): \( W = n \cdot R \cdot \Delta T \)

Substitute the values of \( C_p \) and \( C_v \):

• \( Q = n \cdot \frac{7}{2} R \cdot \Delta T \)
• \( \Delta U = n \cdot \frac{5}{2} R \cdot \Delta T \)
• \( W = n \cdot R \cdot \Delta T \)

Now, we compute the ratio:

• \( Q : \Delta U : W = \frac{7}{2} nR\Delta T : \frac{5}{2} nR\Delta T : nR\Delta T \)

Simplify the ratio:

• \( Q : \Delta U : W = 7 : 5 : 2 \)

Thus, the correct answer is:

7 : 5 : 2

This option corresponds to the provided correct answer, verifying our calculation.