Question

In an isothermal and reversible process, \(1.6 \times 10^{-2}\) kg O\(_2\) expands from 10 dm\(^3\) to 100 dm\(^3\) at 300 K, work done in the process is (R = 8.314 J/mol·K)

• A. -1436 J
• B. -5744 J
• C. -4308 J
• D. -2872 J

Hint:

In isothermal processes, the work done is related to the volume change, temperature, and number of moles of gas.

Answer 1

The Correct Option is D

Step 1: Applying the formula for work in an isothermal process.
For an isothermal expansion of an ideal gas, the work done \( W \) is given by: \[ W = -nRT \ln \left( \frac{V_2}{V_1} \right) \] Where \( n \) is the number of moles of gas, \( R \) is the gas constant, \( T \) is the temperature, \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively.

Step 2: Substituting the values.
Given: \[ n = \frac{1.6 \times 10^{-2}}{32} = 5 \times 10^{-4} \, \text{mol} \quad \text{(molar mass of O\(_2\) is 32 g/mol)} \] \[ V_1 = 10 \, \text{dm}^3, V_2 = 100 \, \text{dm}^3, R = 8.314 \, \text{J/mol·K} \] \[ W = -5 \times 10^{-4} \times 8.314 \times 300 \ln \left( \frac{100}{10} \right) \] \[ W = -2872 \, \text{J} \]
Step 3: Conclusion.
The correct answer is (D) -2872 J, which is the work done in the process.