NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
A sample of gas is compressed by an average pressure of 0 .50 atmosphere so as to decrease its volume from 400 c m 3 to 200 cm 3 . During the process 8 .00 J of heat flows out to surroundings. The change in internal energy of the system is
Options
- A+ 2.13 J
- B+ 10.13 J
- C– 2.13 J
- D– 10.13 J
Correct answer
A. + 2.13 J
Step-by-step solution
Here, Δ V = 200 - 400 = - 200 c m 3 A we know 1 Litre = 1000 cm 3 ⇒ ΔV = − 0.2 Litre . External Pressure P = 0.50 then Work done = - P Δ V = + 0.50 0.2 = + 0.1 atm Litre and 1 litre-atmosphere = 101 .3 Joule ∴ W = + 10.13 Joule According to First law of thermodynamics q = Δ E - W then Δ E = q + W Given q = - 8.00 J ⇒ ΔE = − 8 + 10.13 = 2.13 Joule