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JEE Advanced Physics Thermodynamics 2024 JEE Advanced 2024 (Paper 1)

JEE Advanced Physics Question (2024) — Solution

Question

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 is the gas constant.] \( array l|l List-I & List-II \\ (P) Work done in the complete cyclic process & (1) R T_0-4 R T_0 2 \\ (Q) Change in the internal energy of the gas in the process JK & (2) 0 \\ (R) Heat given to the gas in the process KL & (3) 3 R T_0 \\ (S) Change in the internal energy of the gas in the process MJ & (4) -2 R T_0 2 \\ & (5) -3 R T_0 2 array \)

Options

  1. A. P 1 ; Q 3 ; R 5 ; S 4
  2. B. P 4 ; Q 3 ; R 5 ; S 2
  3. C. P 4; Q 1 ; R 2 ; S 2
  4. D. P 2 ; Q 5 ; R 3 ; S 4

Answer

B. P 4 ; Q 3 ; R 5 ; S 2

Step-by-step solution

aligned & J ( P _0, ~V _0, ~T _0 ) \\ & K ( P _0, 3 ~V _0, 3 ~T _0 ) \\ & M (2 P _0, V _0 2 , ~T _0 ) \\ & L (2 P _0, 3 ~V _0 2 , 3 ~T _0 ) \\ & P _0 ~V _0= nRT _0 \\ & JK isobaric W = P _0 (2 ~V _0 )=2 nRT _0 \\ & U = 3 2 nR (2 ~T _0 )=3 nRT _0 \\ & KL isothermal W = nR (3 ~T ) ( 1 2 )=-3 nRT _0 n 2 \\ & U =0 Q =-3 nRT _0 2 \\ & LM isobaric =2 P _0 (- V _0 )=-2 nRT e _0 \\ & MJ isothermal nRT n _0 2 ; U =0 \\ & WD net =-2 nRT e _0 2 \\ & P 4, Q 3, R 5, ~S 2 aligned

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Related: Physics — Thermodynamics · All PYQ Banks