Skip to content
dronacharya.ai
← JEE Advanced 2024 · Paper 1
PhysicsList match+3 / -1Question 31 of 51

JEE Advanced 2024 · Paper 1, Physics question 14

One mole of a monatomic ideal gas undergoes the cyclic process J→K→L→M→J, as shown in the P-T diagram.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

[R\mathcal{R} is the gas constant.]

P-T diagram: pressure P on the vertical axis, temperature T on the horizontal axis. The cycle is a rectangle with corners J $(T_0, P_0)$, K $(3T_0, P_0)$, L $(3T_0, 2P_0)$ and M $(T_0, 2P_0)$, traversed J→K (isobaric heating at $P_0$, arrow pointing right), K→L (isothermal at $3T_0$, pressure increasing, arrow up), L→M (isobaric cooling at $2P_0$, arrow left), M→J (isothermal at $T_0$, pressure decreasing, arrow down). Dashed guide lines mark $P_0$, $2P_0$ on the P axis and $T_0$, $3T_0$ on the T axis.

List-I

  • P
    Work done in the complete cyclic process
  • Q
    Change in the internal energy of the gas in the process JK
  • R
    Heat given to the gas in the process KL
  • S
    Change in the internal energy of the gas in the process MJ

List-II

  • 1
    RT0−4RT0ln⁡2\mathcal{R}T_0 - 4\mathcal{R}T_0\ln 2
  • 2
    00
  • 3
    3RT03\mathcal{R}T_0
  • 4
    −2RT0ln⁡2-2\mathcal{R}T_0\ln 2
  • 5
    −3RT0ln⁡2-3\mathcal{R}T_0\ln 2
Choose one option
Answer and worked solution

Official answer: B

One mole of monatomic gas, CV=32RC_V = \frac32 R.

Process Type WW by gas ΔU\Delta U
JK isobaric, T0→3T0T_0\to3T_0 2RT02RT_0 3RT03RT_0
KL isothermal at 3T03T_0, P0→2P0P_0\to2P_0 3RT0ln⁡123RT_0\ln\frac12 00
LM isobaric, 3T0→T03T_0\to T_0 −2RT0-2RT_0 −3RT0-3RT_0
MJ isothermal at T0T_0, 2P0→P02P_0\to P_0 RT0ln⁡2RT_0\ln 2 00
  • (P) Wcycle=−3RT0ln⁡2+RT0ln⁡2=−2RT0ln⁡2W_{\text{cycle}} = -3RT_0\ln2 + RT_0\ln2 = -2RT_0\ln2 → (4)
  • (Q) ΔUJK=3RT0\Delta U_{JK} = 3RT_0 → (3)
  • (R) QKL=WKL=−3RT0ln⁡2Q_{KL} = W_{KL} = -3RT_0\ln2 → (5)
  • (S) ΔUMJ=0\Delta U_{MJ} = 0 → (2)

Option (B).

Topics:Thermodynamic processesFirst law and internal energyWork done by a gas
← Previous questionNext question →