Question

Two ideal gases A and B of the same number of moles expand at constant temperatures \( T_1 \) and \( T_2 \) respectively such that the pressure of gas A decreases by \( 50% \) and the pressure of gas B decreases by \( 75% \). If the work done by both the gases is the same, then the ratio \( T_1:T_2 \) is:

• A. \(1:3\)
• B. \(2:3\)
• C. \(3:4\)
• D. \(2:1\)

Hint:

For isothermal expansion of an ideal gas, the work done is given by \( W = nRT \ln \left(\frac{P_i}{P_f} \right) \). When comparing two gases under similar expansion conditions, use logarithmic properties to simplify calculations.

Answer 1

The Correct Option is D

Step 1: Work done during isothermal expansion
For an ideal gas undergoing isothermal expansion, the work done is given by: \[ W = nRT \ln \left(\frac{P_i}{P_f} \right) \] where \( n \) is the number of moles, \( R \) is the universal gas constant, \( T \) is the temperature, and \( P_i \) and \( P_f \) are the initial and final pressures, respectively. Step 2: Work done for gases A and B For gas A: \[ \frac{P_f}{P_i} = \frac{50}{100} = 0.5 \] \[ W_A = nRT_1 \ln \left(\frac{1}{0.5} \right) = nRT_1 \ln 2 \] For gas B: \[ \frac{P_f}{P_i} = \frac{25}{100} = 0.25 \] \[ W_B = nRT_2 \ln \left(\frac{1}{0.25} \right) = nRT_2 \ln 4 \] Step 3: Equating the work done for both gases \[ nRT_1 \ln 2 = nRT_2 \ln 4 \] \[ T_1 \ln 2 = T_2 \ln 4 \] Since \( \ln 4 = 2\ln 2 \), we can rewrite the equation as: \[ T_1 \ln 2 = 2T_2 \ln 2 \] Dividing by \( \ln 2 \), we get: \[ T_1 = 2T_2 \] Step 4: Conclusion \[ T_1:T_2 = 2:1 \] Thus, the correct answer is option (D) 2:1.