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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2018 · Paper 2 · Physics · Thermodynamics

One mole of a monatomic ideal gas undergoes four thermodynamic...

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV - diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-1 with the corresponding statements in List-II.
One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV - diagram

 LIST -I LIST -II
A)In process IIP)Work done by the gas is zero
B)In process IIIIQ)Temperature of the gas remains unchanged
C)In process IIIIIIR)No heat is exchanged between the gas and its surroundings
D)In process IVIVS)Work done by the gas is 6P0V06{P}_{0}{V}_{0}

Options

  1. Aa-q;b-r;c-s;d-p;
  2. Ba-r;b-p;c-s;d-q;
  3. Ca-r;b-s;c-p;d-q;
  4. Da-p;b-s;c-q;d-r;

Correct answer

C. a-r;b-s;c-p;d-q;

Step-by-step solution

Process - II is an adiabatic process
∆Q=∆U+W∆ Q = ∆ U + W
∆Q=0∆ Q = 0
W=−∆UW = - ∆ U
Volume of gas is decreasing ⇒W<0⇒ W < 0
∆U>0∆ U > 0
⇒⇒ Temperatuer of gas increases
⇒⇒ No heat is exchanged between the gas and surrounding.
Process - IIII is an isobaric process
(Pressure remain constant)
W=P∆V=3P0[3V0−V0]=6P0V0W = P ∆ V = 3{P}_{0}\left[3{V}_{0} - {V}_{0}\right] = 6{P}_{0}{V}_{0}
Process - IIIIII is an isochoric process
(Volume remain constant)
∆Q=∆U+W∆ Q = ∆ U + W
W=0W = 0
∆Q=∆U∆ Q = ∆ U
Process - IVIV is an isothermal process
(Temperature remains constant)
∆Q=∆U+W∆ Q = ∆ U + W
∆U=0∆ U = 0
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