Question

If \( \Delta Q \) is the amount of heat supplied to \( n \) moles of a diatomic gas at constant pressure, \( \Delta U \) is the change in internal energy and \( \Delta W \) is the work done, then \( \Delta W : \Delta U : \Delta Q \) is

• A. 2:3:4
• B. 1:2:3
• C. 2:5:7
• D. 5:7:9

Hint:

In thermodynamic processes, the ratio of heat, work, and change in internal energy can often be determined using the specific heats of the gas. For diatomic gases, remember the ratio \( C_p : C_v = 7 : 5 \).

Answer 1

The Correct Option is C

Step 1: First law of thermodynamics.
From the first law of thermodynamics, we know: \[ \Delta Q = \Delta U + \Delta W \] For a diatomic gas at constant pressure: \[ \Delta Q = n C_p \Delta T \] \[ \Delta U = n C_v \Delta T \] \[ \Delta W = p \Delta V = n R \Delta T \] Where: - \( C_p \) and \( C_v \) are the specific heats at constant pressure and volume, respectively. - For a diatomic gas, \( C_p = \frac{7}{5} R \) and \( C_v = \frac{5}{3} R \). Step 2: Applying the relation.
We know that: \[ \Delta W : \Delta U : \Delta Q = 2 : 5 : 7 \] Thus, the correct answer is (C) 2:5:7.