Question

2000 mmol of an ideal gas expanded isothermally and reversibly from 20 L to 30 L at 300 K, calculate the work done in the process (R = 8.314 J K$^{-1}$ mol$^{-1}$).

Hint:

For isothermal and reversible processes, the work done is dependent on the volume change and temperature. Always ensure that the temperatures and volumes are in the correct units for accurate calculation.

Answer 1

For an isothermal and reversible expansion, the work done by an ideal gas is given by the equation: \[ W = -nRT \ln \left( \frac{V_f}{V_i} \right) \] Where: - \(n\) is the number of moles of gas, - \(R\) is the universal gas constant, - \(T\) is the temperature, - \(V_f\) and \(V_i\) are the final and initial volumes, respectively. Step 1: Convert moles of gas to mols.
Given that the number of moles is 2000 mmol = 2 mol, the formula becomes: \[ W = -2 \times 8.314 \times 300 \ln \left( \frac{30}{20} \right) \] Step 2: Simplify the equation.
\[ W = -2 \times 8.314 \times 300 \ln(1.5) \] Step 3: Calculate the work done.
\[ W = -2 \times 8.314 \times 300 \times 0.4055 \] \[ W = -2014.5 \, \text{J} \] Thus, the work done is \(-2014.5 \, \text{J}\). The negative sign indicates work is done by the gas.