Question

A gas of 0.5 mole at 300 K expands isothermally from an initial volume of 2.0 litre to a final volume of 6.0 litre. What is the work done by gas?

Hint:

For {isothermal} processes (constant temperature), the work done involves a natural logarithm (\(\ln\)) of the volume ratio. For {isobaric} processes (constant pressure), work is simply \(P \Delta V\).

Answer 1

The work done (\(W\)) by a gas during an isothermal expansion is given by the formula: \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \] Given:
Number of moles, \(n = 0.5\)
Universal gas constant, \(R \approx 8.314 \, J/mol \cdot K\)
Temperature, \(T = 300 \, K\)
Initial volume, \(V_i = 2.0 \, L\)
Final volume, \(V_f = 6.0 \, L\)
Substitute the values into the formula: \[ W = 0.5 \times 8.314 \times 300 \times \ln\left(\frac{6.0}{2.0}\right) \] \[ W = 0.5 \times 8.314 \times 300 \times \ln(3) \] Using the value \(\ln(3) \approx 1.0986\): \[ W = 1247.1 \times 1.0986 \] \[ W \approx 1370.16 \, J \] The work done by the gas is approximately 1370 J.