Question

2 moles of a monatomic gas requires heat energy \( Q \) to be heated from 30°C to 40°C at constant volume. The heat energy required to raise the temperature of 4 moles of a diatomic gas from 28°C to 33°C at constant volume is:

• A. \( 2Q \)
• B. \( \frac{7Q}{2} \)
• C. \( \frac{4Q}{3} \)
• D. \( \frac{5Q}{3} \)

Hint:

Remember that the specific heat capacity for diatomic gases is higher than for monatomic gases, which affects the energy required for the same temperature change.

Answer 1

The Correct Option is D

For an ideal gas, the heat energy required to raise the temperature can be expressed using the equation: \[ Q = n C_V \Delta T \] Where: - \( n \) is the number of moles, - \( C_V \) is the specific heat at constant volume, - \( \Delta T \) is the change in temperature. For a monatomic gas: \[ Q_{\text{monatomic}} = n C_V \Delta T \] For a diatomic gas, the specific heat at constant volume \( C_V \) is higher. For monatomic gas \( C_V = \frac{3}{2} R \) and for diatomic gas \( C_V = \frac{5}{2} R \). The heat energy required for 2 moles of a monatomic gas to go from 30°C to 40°C is \( Q \), so: \[ Q = 2 \times \frac{3}{2} R \times 10 = 30 R \] For 4 moles of a diatomic gas, the temperature change is from 28°C to 33°C, so: \[ Q_{\text{diatomic}} = 4 \times \frac{5}{2} R \times 5 = 50 R \] Since the heat energy for the monatomic gas was \( 30 R \), we find that: \[ Q_{\text{diatomic}} = \frac{5Q}{3} \] Thus, the correct answer is \( \frac{5Q}{3} \).