| ⇦ |
| ⇨ |
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: - The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV
- The wetability of a surface by a liquid depends primarily on
- Amout of heat required to raise the temperature of a body through 1 K is called its
- The horizontal range and maximum height attained by a projectile are R and H,respectively
- Copper and carbon wires are connected in series and the combined resistor is kept
Topics: Behavior of Perfect Gas and Kinetic Theory
(34)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV
- The wetability of a surface by a liquid depends primarily on
- Amout of heat required to raise the temperature of a body through 1 K is called its
- The horizontal range and maximum height attained by a projectile are R and H,respectively
- Copper and carbon wires are connected in series and the combined resistor is kept
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 🙂