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JEE Advanced 2014Paper 2
Exam
JEE Advanced
Year
2014
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2014 · Paper 2 · Physics · Thermodynamics

Paragraph: In the figure a container is shown to have a movable...

Paragraph: In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 22 moles of an ideal monatomic gas at 700K700 K and the upper compartment is filled with 22 moles of an ideal diatomic gas at 400K400 K. The heat capacities per mole of an ideal monatomic gas are CV=32R,CP=52RC_{V}=\frac{3}{2} R, C_{P}=\frac{5}{2} R, and those for an ideal diatomic gas are CV=52R,CP=72RC_{V}=\frac{5}{2} R, C_{P}=\frac{7}{2} R. Paragraph: In the figure a container is shown to have a movable (without friction) piston on top. The container and the

Question: Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

Options

  1. A250 R
  2. B200 R
  3. C100 R
  4. D- 100 R

Correct answer

D. - 100 R

Step-by-step solution

Heat given by lower compartment =2×52R×(700−T)= 2 \times \dfrac{5}{2}R \times \left(700 - T\right) ...(i)
Heat obtained by upper compartment =2×72R×(T−400)= 2 \times \dfrac{7}{2}R \times \left(T - 400\right) ...(ii)
By equating (i) and (ii)
5(700−T)=7(T−400)5\left(700 - T\right) = 7\left(T - 400\right)
3000−5T=7T−28003000 - 5T = 7T - 2800
6300=12T6300 = 12T
T=525KT = 525K
∴∴ Work done by lower gas =nR∆T=−350R= nR ∆ T = - 350R
Work done by upper gas =nR∆T=+250R= nR ∆ T = + 250R
Net work done −100R- 100R
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