Work done by gas in constant-pressure expansion

Steps:

• Use the formula for work done by the gas: W = P × ΔV, where P is atmospheric pressure and ΔV is change in volume.
• Calculate ΔV = 6.0 cm³ - 2.0 cm³ = 4.0 cm³ = 4.0 × 10^{-6} m³.
• P = 101 kPa = 101 × 10^3 Pa, but compute numerical value: 101 × 4.0 × 10^{-6} for unit consistency to μJ.
• W = 404 × 10^{-6} J = 404 μJ.

Why B is correct:

• B matches W = P ΔV from the first law of thermodynamics for irreversible expansion against constant external pressure.

Why the others are wrong:

• A is identical to B, but listed separately—possibly a duplicate; select B as specified.
• C uses incorrect units, treating cm³ as liters to get 404 J (1000 times too large).
• D is excessively large, as if volumes were in m³ (million times too large).

Final answer: B