JEE Main 2020 · 4 September · evening · Single correct · +4 / −1
Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume. The work done is :
- (a) Zero
- (b) -RT ln V₂/V₁
- (c) C V (T₂ – T₁)
- (d) – RT (V₂ – V₁)
Answer and solution
Answer (a) Zero
Solution
Chemistry ma’am · 3 steps
Expanding into vacuum means nothing pushes back, so the gas does no work at all.
- Work needs resistance. Expansion work is done against the external pressure, not the gas's own pressure. So it is w = -∫ pₑₓₜ dV.
- Vacuum outside. In a vacuum there is nothing to push against, so pₑₓₜ = 0 throughout the expansion. This is called free expansion.
w = -pₑₓₜ Δ V = -(0)×(V₂ - V₁) = 0
- So. The work done is zero, option (a). The 5 moles, 1 bar, 298 K and the doubling of volume do not matter here, because pₑₓₜ = 0 makes the whole product zero.
The idea to keep. Always ask what pressure the gas is pushing against. Work uses pₑₓₜ, and for expansion into vacuum pₑₓₜ = 0, so w = 0 however much the volume grows.
Answer from the NTA final answer key. Solution by PrepWiser.