| ⇦ |
| ⇨ |
40 g of Argon is heated from 40⁰C to 100⁰C (R=2 cal/mole). Heat absorbed at constant volume is
Options
(a) 100 cal
(b) 80 cal
(c) 180 cal
(d) 120 cal
Correct Answer:
180 cal
Explanation:
Heat absorbed at constant volume = nCvdT
Now argon is monoatomic Cv = (3/2) R
Number of moles = 4/0/40 = 1
.·. Q = 1 × (3/2) × 2 × (100-40) = 3 × 60 = 180 cal.
Related Questions: - Two capacitors having capacitances C₁ and C₂ are charged with 120V and 200V batteries
- Surface tension of water will be minimum at
- Two closed pipes produce 10 beats per second when emitting their fundamental nodes.
- A liquid wets a solid completely. The meniscus of the liquid in a sufficiently long
- A small object of uniform density rolls up a curved surface with an initial velocity
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two capacitors having capacitances C₁ and C₂ are charged with 120V and 200V batteries
- Surface tension of water will be minimum at
- Two closed pipes produce 10 beats per second when emitting their fundamental nodes.
- A liquid wets a solid completely. The meniscus of the liquid in a sufficiently long
- A small object of uniform density rolls up a curved surface with an initial velocity
Topics: Thermodynamics (179)
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

Leave a Reply