Question

A gas with a ratio of specific heats, \( \gamma = \frac{4}{3} \), is heated isobarically. The percentage of the given heat used in external work done is

• A. 15 %
• B. 25 %
• C. 50 %
• D. 60 %

Hint:

In thermodynamics, the ratio of specific heats \( \gamma \) is crucial for determining the distribution of heat energy between internal energy and work done in various processes.

Answer 1

The Correct Option is B

Given: - Ratio of specific heats, \( \gamma = \frac{4}{3} \) Step 1: Determine the fraction of heat used for work in an isobaric process For an isobaric process, the heat \( Q \) added to the gas is used to increase the internal energy \( \Delta U \) and to do work \( W \). 
The relationship is given by: \[ Q = \Delta U + W \] For an ideal gas, the work done \( W \) in an isobaric process is: \[ W = P \Delta V = n R \Delta T \] The change in internal energy \( \Delta U \) is: \[ \Delta U = n C_v \Delta T \] The total heat added \( Q \) is: \[ Q = n C_p \Delta T \] The fraction of heat used for work is: \[ \frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p} \] Given that \( \gamma = \frac{C_p}{C_v} \) and \( C_p = C_v + R \), we can express \( C_p \) as: \[ C_p = \frac{\gamma R}{\gamma - 1} \] Thus, \[ \frac{W}{Q} = \frac{R}{\frac{\gamma R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma} \] Substituting \( \gamma = \frac{4}{3} \): \[ \frac{W}{Q} = \frac{\frac{4}{3} - 1}{\frac{4}{3}} = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{4} = 0.25 \] Step 2: Convert the fraction to a percentage \[ 0.25 \times 100 = 25 % \] Final Answer: 25 %