| ⇦ | 
| ⇨ |
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: - Four identical rods are joined to from a square. If the temperature difference
- A neutron is moving with a velocity u.It collides head on and elastically with an atom
- In the Davisson and Germer experiment, the velocity of electrons emitted from the
- If n₁, n₂ and n₃ are the fundamental frequencies of three segments into which a string
- A particle doing SHM has amplitude=4 cm and time period = 12 sec. The ratio between times
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Four identical rods are joined to from a square. If the temperature difference
- A neutron is moving with a velocity u.It collides head on and elastically with an atom
- In the Davisson and Germer experiment, the velocity of electrons emitted from the
- If n₁, n₂ and n₃ are the fundamental frequencies of three segments into which a string
- A particle doing SHM has amplitude=4 cm and time period = 12 sec. The ratio between times
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