NTA Abhyas JEE Main2020PhysicsThermodynamicsPractice
An ideal monatomic gas is confined in a cylinder by a spring-loaded massless piston of cross-section 8 × 10 − 3 m 2 . The piston can slide on the walls of the cylinder without any friction. Initially, the gas is at 300 K and occupies a volume of 2 .4 × 10 − 3 m 3 and the spring is in its relaxed position. Now, gas is slowly heated by a small electric heater and the piston moves out slowly by 0 .1 m . Calculate the fi
Options
- AT 2 = 600   K ,   Q = 680   J
- BT 2 = 800   K ,   Q = 600   J
- CT 2 = 600   K ,   Q = 720   J
- DT 2 = 800   K ,   Q = 720   J
Correct answer
D. T 2 = 800   K ,   Q = 720   J
Step-by-step solution
In the initial condition P 1 = 10 5   Pa , V 1 = 2 . 4 × 10 - 3   m 3 , T 1 = 300   K Finally, P 2 = P 1 + k x A = 2 × 10 5   Pa V 2 = V 1 + A x = 3 . 2 × 10 - 3   m 3 P 1 V 1 T 1 = P 2 V 2 T 2 , on solving we get T 2 = 800   K Applying work-energy theorem for the piston W gas + W atmos + W spring = 0 W gas = P 0 A x + 1 2 k x 2 = 120   J Δ U = n C V Δ T = n R Δ T γ - 1 = P 2 V 2 - P 1 V 1 γ - 1 = 600   J Q = Δ U + W g a s = 6 0 0