| ⇦ |
| ⇨ |
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure.

The change in internal energy of the gas during the transition is
Options
(a) -20 kJ
(b) 20 J
(c) -12 kJ
(d) 20 kJ
Correct Answer:
-20 kJ
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - An electron moves on a straight line path XY as shown. The abcd is a coil adjacent
- A body of mass 1.0 kg is falling with an acceleration of 10 ms⁻².
- An object is placed 30 cm away from a convex lens of focal length 10 cm and a sharp
- 540 g of ice at 0° C is mixed with 540 g of water at 80° C. The final temperature
- The value of Poisson’s ratio lies between
Topics: Behavior of Perfect Gas and Kinetic Theory
(34)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An electron moves on a straight line path XY as shown. The abcd is a coil adjacent
- A body of mass 1.0 kg is falling with an acceleration of 10 ms⁻².
- An object is placed 30 cm away from a convex lens of focal length 10 cm and a sharp
- 540 g of ice at 0° C is mixed with 540 g of water at 80° C. The final temperature
- The value of Poisson’s ratio lies between
Topics: Behavior of Perfect Gas and Kinetic Theory (34)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

ΔU = nCΔT
Also, T = PV/nR
Now,
ΔT = T-T => [PV – PV] / nR
ΔU = nR/¥-1 {PV- PV / nR} [here, ¥ = gamma]
(nR will be cancelled out)
We get,
{-8 x 10^3} / {2/5} = -20kJ.
Hope it helps 🙂