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JEE Advanced · Physics · 13. Thermodynamics

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure).Initially the gas is at temperature T1 , pressure P1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2 , pressure P2 and volume V2 . During this process the piston moves out by a distance x . Ignoring the friction between the piston the cylinder, the correct statement(s) is (are)

  1. A If V2=2V1 and T2=3T1 , then the energy stored in the spring is 14P1V1
  2. B If V2=2V1 and T2=3T1 , then the change in internal energy is 3P1V1
  3. C If V2=3V1 and T2=4T1 , then the work done by the gas is 73P1V1
  4. D If V2=3V1 and T2=4T1 , then the heat supplied to the gas is 176P1V1
Verified Solution

Answer & Solution

Correct Answer

(A) If V2=2V1 and T2=3T1 , then the energy stored in the spring is 14P1V1

Step-by-step Solution

Detailed explanation



(i) P=P1+KxA
P2=32P1x=V1A
3P12=P1+KxA
Kx=P1A2
Energy of spring
12Kx2=P1A4x=P1V14
(ii) U=f2P2V2-P2V1
=3P1V1
(iii) Pf=4P13
KX=P13A
X=2V1A
Wgas=-WPatm+Wspring
=P1Ax+12Kx.x
=+P1A.2V1A+12.P1A3.2V1A
=2P1V1+P1V13=7P1V13
(iv) Q=W+U
=7P1V13+32P2V2-P1V1
=7P1V13+3243P1.3V1-P1V1
=7P1V13+92P1V1=41P1V16
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