| ⇦ |
| ⇨ |
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: - A particle moves from position r₁ = 3i + 2j -6k to position r₂ = 14i + 13 j+ 9 k
- A particle of mass m, charge Q and kinetic enery T enters a transverse
- The pressure at the bottom of a tank containing a liquid does not depend on
- A vibrating string of certain length l under a tension T resonates with a mode
- Two bulbs 60 W and 100 W designed for voltage 220 V are connected in series
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle moves from position r₁ = 3i + 2j -6k to position r₂ = 14i + 13 j+ 9 k
- A particle of mass m, charge Q and kinetic enery T enters a transverse
- The pressure at the bottom of a tank containing a liquid does not depend on
- A vibrating string of certain length l under a tension T resonates with a mode
- Two bulbs 60 W and 100 W designed for voltage 220 V are connected in series
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