Question

When 2 moles of a monatomic gas expands adiabatically from a temperature of 80°C to 50°C, the work done is \( W \). The work done when 3 moles of a diatomic gas expands adiabatically from 50°C to 20°C is:

• A. 7 W
• B. 5 W
• C. 2.5 W
• D. 3.5 W

Hint:

For adiabatic expansion problems, use the molar heat capacities and the formula \( W = n C_V \Delta T \) to calculate the work done.

Answer 1

The Correct Option is C

Given: Monatomic gas: 2 moles, expanding adiabatically from \( 80^\circ C \) to \( 50^\circ C \). Diatomic gas: 3 moles, expanding adiabatically from \( 50^\circ C \) to \( 20^\circ C \). 

Step 1: Work done by a gas in an adiabatic process The work done \( W \) in an adiabatic expansion or compression is given by the formula: \[ W= n C_V \frac{\Delta T}{\gamma - 1} \] Where:
\( n \) = number of moles,
\( C_V \) = molar heat capacity at constant volume,
\( \gamma \) = adiabatic index (\( \gamma = \frac{C_P}{C_V} \)),
\( \Delta T \) = temperature change.

Step 2: Work done by the monatomic gas For a monatomic gas, \( \gamma_{\text{monatomic}} = \frac{5}{3} \). For 2 moles of the monatomic gas, the temperature change \( \Delta T_1 = 50^\circ C - 80^\circ C = -30^\circ C \). The work done is: \[ W_{\text{monatomic}} = n_1 C_V \frac{\Delta T_1}{\gamma_{\text{monatomic}} - 1} \] Substitute the values: \[ W_{\text{monatomic}} = 2 C_V \frac{-30}{\frac{5}{3} - 1} = 2 C_V \frac{-30}{\frac{2}{3}} = 2 C_V \times (-45) = -90 C_V \] Thus, the work done by the monatomic gas is \( W_{\text{monatomic}} = -90 C_V \). 

Step 3: Work done by the diatomic gas For a diatomic gas, \( \gamma_{\text{diatomic}} = \frac{7}{5} \). For 3 moles of the diatomic gas, the temperature change \( \Delta T_2 = 20^\circ C - 50^\circ C = -30^\circ C \). The work done is: \[ W_{\text{diatomic}} = n_2 C_V \frac{\Delta T_2}{\gamma_{\text{diatomic}} - 1} \] Substitute the values: \[ W_{\text{diatomic}} = 3 C_V \frac{-30}{\frac{7}{5} - 1} = 3 C_V \frac{-30}{\frac{2}{5}} = 3 C_V \times (-75) = -225 C_V \] Thus, the work done by the diatomic gas is \( W_{\text{diatomic}} = -225 C_V \). 

Step 4: Ratio of work done Now, we find the ratio of the work done by the diatomic gas to the work done by the monatomic gas: \[ \frac{W_{\text{diatomic}}}{W_{\text{monatomic}}} = \frac{-225 C_V}{-90 C_V} = \frac{225}{90} = 2.5 \] 

Thus, the work done by the diatomic gas is \( 2.5 \) times the work done by the monatomic gas.