Question

At \( T = 27^\circ C \), \( V_i = 0.4 \, \text{L} \), \( V_f = 0.8 \, \text{L} \), find isothermal work.

Hint:

The isothermal work is calculated using the formula \( W = nRT \ln \frac{V_f}{V_i} \), where \( n \) is the number of moles and \( T \) is the temperature in Kelvin.

Answer 1

The volume exactly doubles here (0.4 L to 0.8 L), so \( V_f/V_i = 2 \) and \( \ln 2 \approx 0.693 \). With \( T = 300 \, \text{K} \), the isothermal work comes out to \[ W = nRT\ln 2 \approx nR(300)(0.693) \] \[ \boxed{W \approx 208\, nR} \] leaving the final numeric value dependent on the number of moles \( n \) and the gas constant \( R \) used.