PrepRung
Sign inSign up free

JEE Advanced 2023 Paper 1 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

One mole of a monatomic ideal gas is heated through a process in which its temperature rises, so that it absorbs heat Q>0 and does work W. In List-I…

Single correct option

One mole of a monatomic ideal gas is heated through a process in which its temperature rises, so that it absorbs heat Q>0Q>0 and does work WW. In List-I the process is described by its equation; in List-II the value of W/QW/Q is given. Match List-I with List-II.

List-I
(P) Isobaric expansion
(Q) Process PV2=constantPV^{2}=\text{constant}
(R) Process P∝VP\propto V
(S) Process PV1/2=constantPV^{1/2}=\text{constant}

List-II
(1) −2-2
(2) 14\dfrac14
(3) 25\dfrac25
(4) 47\dfrac47
(5) 00
  1. A. P→3, Q→1, R→2, S→4
  2. B. P→3, Q→1, R→4, S→2
  3. C. P→5, Q→1, R→2, S→4
  4. D. P→3, Q→2, R→1, S→4
Show answer and solution

Answer: (A) P→3, Q→1, R→2, S→4

For PVn=PV^n= const and one mole, W=R ΔT1−nW=\dfrac{R\,\Delta T}{1-n} and the molar heat capacity is C=CV+R1−nC=C_V+\dfrac{R}{1-n} with CV=32RC_V=\tfrac32R, so WQ=1/(1−n)3/2+1/(1−n)\dfrac WQ=\dfrac{1/(1-n)}{3/2+1/(1-n)}. (P) Isobaric is n=0n=0: 13/2+1=25\dfrac{1}{3/2+1}=\dfrac25, which is (3). (Q) n=2n=2: W=−RΔTW=-R\Delta T, C=12RC=\tfrac12R, so W/Q=−2W/Q=-2, which is (1) (the gas is compressed while absorbing heat). (R) P∝VP\propto V is n=−1n=-1: 1/23/2+1/2=14\dfrac{1/2}{3/2+1/2}=\dfrac14, which is (2). (S) n=12n=\tfrac12: W=2RΔTW=2R\Delta T, C=32R+2R=72RC=\tfrac32R+2R=\tfrac72R, so W/Q=47W/Q=\dfrac47, which is (4). List-II item (5) is not used. Thus P→3, Q→1, R→2, S→4.
Report a mistake

Original question written in the JEE Advanced 2023 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake