JEE MainPhysicsThermodynamics
A thermodynamic system undergoes an isobaric expansion at a constant pressure of 2 10^5 Pa . During this process, it is observed that 75 % of the total heat supplied to the system is used to increase its internal energy. If the increase in internal energy is 6000 J , the change in volume of the system is :
Options
- A30 L
- B10 L
- C90 L
- D0.01 L
Correct answer
B. 10 L
Step-by-step solution
Given, increase in internal energy, U = 6000 J Since 75 % of the total heat supplied ( Q ) is used to increase the internal energy: U = 0.75 Q Q = 6000 0.75 = 8000 J According to the first law of thermodynamics: Q = U + W W = 8000 - 6000 = 2000 J Work done in an isobaric process is given by W = P V 2000 = (2 10^5) V V = 2000 2 10^5 = 0.01 m ^3 Converting to liters: V = 0.01 1000 L = 10 L Answer: 10 L