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JEE Advanced Physics Thermodynamics 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Physics Question (2025) — Solution

Question

An ideal monatomic gas of n moles is taken through a cycle W X Y Z W consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic V-T diagram. The volume of the gas at W, X and Y points are, 64 ~cm ^3, 125 ~cm ^3 and 250 ~cm ^3, respectively. If the absolute temperature of the gas T_W at the point W is such that n R T_W=1 ~J ( R is the universal gas constant), then the amount of heat absorbed (in J ) by the gas along the path X Y is

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

aligned & nRT _ W = P _ W ~V _ W =1 ~J \\ & P _ W = 1 64 10^6 ~Pa aligned For WX process aligned & P_X V_X^Y=P_W V_W^y \\ & P_X=P_W ( V_W V_X )^y aligned amount of heat absorbed in XY process aligned Q & = nCP T = n 5 2 R [ T _ Y - T _ X ] [ For monoatomic gas C _ P = 5 R 2 ] \\ Q & = 5 2 [ nRT _ Y - nRT T _ X ] \\ & = 5 2 [ P _ Y V _ Y - P _ X V _ X ] \\ & = 5 2 P _ X [ V _ Y - V _ X ] [ P _ X = P _ Y ; Isobaric process ] \\ & = 5 2 P _ W [ V _ W V _ X ]^ y [ V _ Y - V _ X ] aligned Putting values : Q=1.6 Joule

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