| ⇦ | 
| ⇨ |
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 liquid wets a solid completely. The meniscus of the liquid in a sufficiently long
- which of the following can not be emitted by radioactive substances during their decay?
- The kirchoff’s first law (∑i=0) and second law (∑iR=∑E), where the symbols
- In adiabatic expansion, product of PV
- Carbon,silicon and germanium have four valence electrons each. At room temperature
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A liquid wets a solid completely. The meniscus of the liquid in a sufficiently long
- which of the following can not be emitted by radioactive substances during their decay?
- The kirchoff’s first law (∑i=0) and second law (∑iR=∑E), where the symbols
- In adiabatic expansion, product of PV
- Carbon,silicon and germanium have four valence electrons each. At room temperature
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