Question

The heat energy supplied to a diatomic gas at constant pressure is 210 J, then the work done by the gas is

• A. \( 60 \) J
• B. \( 150 \) J
• C. \( 90 \) J
• D. \( 210 \) J

Hint:

For an isobaric process, the ratio of work done to heat supplied depends on the molar heat capacities. For a diatomic gas, \( C_p = \frac{7}{2} R \). Use the relationship \( W = nR \Delta T \) and \( Q = n C_p \Delta T \) to find the ratio \( \frac{W}{Q} \). Then, calculate the work done using the given heat supplied.

Answer 1

The Correct Option is A

For a diatomic gas at constant pressure (isobaric process), the heat supplied \( Q \) is related to the change in internal energy \( \Delta U \) and the work done by the gas \( W \) by the first law of thermodynamics: \( Q = \Delta U + W \).
For a diatomic ideal gas, the molar heat capacity at constant pressure is \( C_p = \frac{7}{2} R \), and the molar heat capacity at constant volume is \( C_v = \frac{5}{2} R \).
The ratio of specific heats is \( \gamma = \frac{C_p}{C_v} = \frac{7/2 R}{5/2 R} = \frac{7}{5} \).
The heat supplied at constant pressure is \( Q = n C_p \Delta T = n \left( \frac{7}{2} R \right) \Delta T \).
Given \( Q = 210 \) J.
The work done by the gas at constant pressure is \( W = P \Delta V = nR \Delta T \).
We can relate \( W \) to \( Q \) using the ratio: $$ \frac{W}{Q} = \frac{nR \Delta T}{n \left( \frac{7}{2} R \right) \Delta T} = \frac{1}{\frac{7}{2}} = \frac{2}{7} $$ So, \( W = \frac{2}{7} Q \).
Given \( Q = 210 \) J, the work done by the gas is: $$ W = \frac{2}{7} \times 210 = 2 \times 30 = 60 \text{ J} $$ The work done by the gas is 60 J.