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PhysicsMatrix match+8Question 39 of 60

IIT-JEE 2011 · Paper 2, Physics question 39

One mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in the P-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.

P-V diagram: A (3V, 3P) → B (V, 3P) isobaric; B → C (V, P) isochoric; C → D (9V, P) isobaric; D → A along a curve (PV constant).

Column I

  • A
    Process A → B
  • B
    Process B → C
  • C
    Process C → D
  • D
    Process D → A

Column II

  • P
    Internal energy decreases.
  • Q
    Internal energy increases.
  • R
    Heat is lost.
  • S
    Heat is gained.
  • T
    Work is done on the gas.
Mark every match for each entry
PQRST
A
B
C
D
Answer and worked solution

Official answer: A → P, R, T; B → P, R; C → Q, S; D → R, T

Temperatures from T=PVRT=\frac{PV}R, in units of PV/RPV/R: A = 9, B = 3, C = 1, D = 9.

  • A → B: isobaric compression; T falls 9 → 3, so U decreases. Work is done on the gas, and Q=ΔU+Wby<0Q=\Delta U+W_\text{by}<0, so heat is lost. → P, R, T.
  • B → C: isochoric; T falls 3 → 1, so U decreases with no work, and heat is lost. → P, R.
  • C → D: isobaric expansion; T rises 1 → 9, so U increases and heat is gained. → Q, S.
  • D → A: PV=9PVPV=9PV at both ends (an isotherm), so ΔU=0\Delta U=0. The gas is compressed (work done on it), so heat is lost. → R, T.

Answer: A → P, R, T; B → P, R; C → Q, S; D → R, T

Topics:Thermodynamic processesFirst law and internal energy
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