NTA Abhyas JEE Main2020PhysicsThermodynamicsPractice
Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T 0 , while box B contains one mole of helium at temperature 7 3 T 0 . The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then , the final temperature of the gases, T f , in
Options
- AT f = 3 7 T 0
- BT f = 7 3 T 0
- CT f = 3 2 T 0
- DT f = 5 2 T 0
Correct answer
C. T f = 3 2 T 0
Step-by-step solution
Here, change in internal energy of the system is zero, ie, increase in internal energy of one is equal to decrease in internal energy of other. ∆ U A = 1 × 5 R 2 ( T f - T o ) ∆ U B = 1 × 3 R 2 ( T f - 7 3 T 0 ) Now, ∆ U A + ∆ U B = 0 5 R 2 T f - T 0 + 3 R 2 T f - 7 T 0 3 = 0 5 T f - 5 T 0 + 3 T f - 7 T 0 = 0 ⇒ 8 T f = 12 T 0 ⇒ T f = 12 8 T 0 = 3 2 T 0