In an isobaric process, \( 2 \, \text{moles} \) of an ideal gas expand from \( 10 \, \text{L} \) to \(
20 \, \text{L} \) at \( 400 \, \text{K} \). What is the work done by the gas? (Take \( R = 8.3 \,
\text{J mol}^{-1} \text{K}^{-1} \))
Work done: \( W = P \Delta V = \mu R T \left(\frac{\Delta V}{V_1}\right) \), but directly, \( W = \mu R
\Delta T \).
Here, \( \Delta V = 20 - 10 = 10 \, \text{L} \), use \( W = P \Delta V = \mu R T \frac{\Delta V}{V} \),
but since \( P V = \mu R T \), \( W = \mu R T \left(\frac{V_2 - V_1}{V_1}\right) \) simplifies via \( P
\).
Actually, \( W = \mu R T \left(\frac{V_2}{V_1} - 1\right) \), but directly: \( W = P \Delta V \), and \(
P V = \mu R T \).
Correctly, \( W = \mu R \Delta T \), but here \( T \) is constant, so \( W = P \Delta V \), and \( P =
\frac{\mu R T}{V} \).
Better: \( W = \mu R T \left(\frac{V_2 - V_1}{V_1}\right) \), no, simply \( W = P \Delta V \), use \( \mu
R T \).
\( W = 2 \times 8.3 \times 400 \times \frac{10}{10} = 6640 \, \text{J} \) (corrected via \( P \Delta V
\)).
Final: \( W = 2 \times 8.3 \times 400 = 6640 \, \text{J} \) (adjust units if needed, assume \( \Delta V
\) in appropriate form).