JEE Main 2020 · 5 September · morning · Single correct · +4 / −1

Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are labelled as A → B, A → C and A → D. The change in internal energies during these process are taken as E AB , E AC and E AD and the work done as W AB , W AC and W AD .
The correct relation between these parameters are :

JEE Main PYQ figure
  1. (a) E AB < E AC < E AD , W AB > 0, W AC > W AD
  2. (b) E AB = E AC = E AD , W AB > 0, W AC = 0, W AD < 0
  3. (c) E AB > E AC > E AD , W AB < W AC < W AD
  4. (d) E AB = E AC < E AD , W AB > 0, W AC = 0, W AD < 0
Answer and solution

Answer (b) E AB = E AC = E AD , W AB > 0, W AC = 0, W AD < 0

Solution

Physics sir · 2 steps

B, C and D all sit on the same isotherm T₁, so the internal energy change is identical for all three paths.

  1. Internal energy. For an ideal gas, internal energy depends only on temperature: Δ U = nC_VΔ T. Every path starts at A on the T₂ isotherm and ends at B, C or D on the T₁ isotherm, so Δ T = T₁ - T₂ each time. Hence E_(AB) = E_(AC) = E_(AD).

    W = ∫ P dV

    PathWhat happens to VWork by the gas
    A → Bincreases (horizontal, isobaric)W_(AB) > 0
    A → Cstays the same (vertical, isochoric)W_(AC) = 0
    A → Ddecreases (D lies to the left of A)W_(AD) < 0
  2. So. Equal internal energy changes, with W_(AB) > 0, W_(AC) = 0 and W_(AD) < 0. That is option (b).

The idea to keep. For an ideal gas, check which isotherms a path starts and ends on: same end temperatures mean the same Δ U, whatever the path. For work, just watch whether the volume grows, stays put or shrinks.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Thermodynamics

Asked about 150 times in JEE Main since 2020. This question is from 2020.