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← IIT-JEE 2008 · Paper 2
PhysicsMatrix match+6Question 44 of 66

IIT-JEE 2008 · Paper 2, Physics question 44

Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.

Column I

  • A
    An insulated container has two chambers separated by a valve. Chamber I contains an ideal gas and the Chamber II has vacuum. The valve is opened.
    An insulated box divided into chamber I (ideal gas, dots) and chamber II (vacuum) joined through a valve in the partition.
  • B
    An ideal monoatomic gas expands to twice its original volume such that its pressure P∝1V2P\propto\dfrac1{V^2}, where VV is the volume of the gas
  • C
    An ideal monoatomic gas expands to twice its original volume such that its pressure P∝1V4/3P\propto\dfrac1{V^{4/3}}, where VV is its volume
  • D
    An ideal monoatomic gas expands such that its pressure PP and volume VV follows the behaviour shown in the graph
    P against V: from V₁ to 2V₁ the pressure rises gradually and then more steeply as volume increases (an expansion with increasing pressure).

Column II

  • P
    The temperature of the gas decreases
  • Q
    The temperature of the gas increases or remains constant
  • R
    The gas loses heat
  • S
    The gas gains heat
Mark every match for each entry
PQRS
A
B
C
D
Answer and worked solution

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

  • A, free expansion into vacuum. No heat (q=0q=0) and no work, so ΔU=0\Delta U=0 and the temperature of an ideal gas stays the same. → Q.
  • B, PV2PV^2 constant. T∝PV∝1VT\propto PV\propto\frac1V: the gas cools. For the polytropic index n=2n=2, the molar heat capacity is C=CV+R1−n=32R−R=R2>0C=C_V+\frac R{1-n}=\frac32R-R=\frac R2>0, and with ΔT<0\Delta T<0 the gas loses heat. → P, R.
  • C, PV4/3PV^{4/3} constant. T∝V−1/3T\propto V^{-1/3}: cools. Here C=32R+R1−4/3=−32R<0C=\frac32R+\frac R{1-4/3}=-\frac32R<0, so with ΔT<0\Delta T<0 the gas gains heat. → P, S.
  • D, P and V both increase. T∝PVT\propto PV rises, and W>0W>0, ΔU>0\Delta U>0, so q=ΔU+W>0q=\Delta U+W>0. → Q, S.

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