| ⇦ | 
| ⇨ |
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: - Two particles of masses m₁ m₂ move with initial velocities u₁ and u₂
- The potential differences across the resistance, capacitance and inductance
- Two parallel metal plates having charges +Q and -Q face each other at a certain
- For a thermocouple, the neutral temperature is 270⁰C when its cold junction is at 20⁰C.
- When a spring is extended by 2 cm energy stored is 100 J. When extended further by 2 cm
Topics: Behavior of Perfect Gas and Kinetic Theory
(34)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two particles of masses m₁ m₂ move with initial velocities u₁ and u₂
- The potential differences across the resistance, capacitance and inductance
- Two parallel metal plates having charges +Q and -Q face each other at a certain
- For a thermocouple, the neutral temperature is 270⁰C when its cold junction is at 20⁰C.
- When a spring is extended by 2 cm energy stored is 100 J. When extended further by 2 cm
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 🙂