JEE Advanced2015PhysicsThermodynamicsActual
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 T 1 , pressure P 1 and volume V 1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T 2 , pressure P 2 and volume V 2 . During this process the piston moves out by a distance x . Ignoring the friction between the piston the cylinder
Options
- AIf V 2 = 2 V 1 and T 2 = 3 T 1 , then the energy stored in the spring is 1 4 P 1 V 1
- BIf V 2 = 2 V 1 and T 2 = 3 T 1 , then the change in internal energy is 3 P 1 V 1
- CIf V 2 = 3 V 1 and T 2 = 4 T 1 , then the work done by the gas is 7 3 P 1 V 1
- DIf V 2 = 3 V 1 and T 2 = 4 T 1 , then the heat supplied to the gas is 17 6 P 1 V 1
Correct answer
A. If V 2 = 2 V 1 and T 2 = 3 T 1 , then the energy stored in the spring is 1 4 P 1 V 1
Step-by-step solution
(i) P = P 1 + K x A P 2 = 3 2 P 1 ⇒ x = V 1 A 3 P 1 2 = P 1 + K x A K x = P 1 A 2 Energy of spring 1 2 K x 2 = P 1 A 4 x = P 1 V 1 4 (ii) ∆ U = f 2 P 2 V 2 - P 2 V 1 = 3 P 1 V 1 (iii) P f = 4 P 1 3 K X = P 1 3 A X = 2 V 1 A W g a s = - W P a t m + W s p r i n g = P 1 A x + 1 2 K x . x = + P 1 A . 2 V 1 A + 1 2 . P 1 A 3 . 2 V 1 A = 2 P 1 V 1 + P 1 V 1 3 = 7 P 1 V 1 3 (iv) ∆ Q = W + ∆ U = 7 P 1 V 1 3 + 3 2 P 2 V 2 - P 1 V 1 = 7 P 1 V 1 3 + 3 2 4 3 P 1 . 3 V 1 - P 1 V 1 = 7 P 1 V 1 3 + 9 2 P 1 V 1 = 41 P 1 V 1 6