Thermodynamics Question 121

Question: In thermodynamic process, 200 Joules of heat is given to a gas and 100 Joules of work is also done on it. The change in internal energy of the gas is

[AMU (Engg.) 1999]

Options:

A) 100 J

B) 300 J

C) 419 J

D) 24 J

Show Answer

Answer:

Correct Answer: B

Solution:

$ \Delta Q=\Delta U+\Delta W $ ;

$ \Delta Q=200J $ and $ \Delta W=-100J $

Therefore $ \Delta U=\Delta Q-\Delta W=200-(-100)=300J $



NCERT Chapter Video Solution

Dual Pane