| ⇦ | 
| ⇨ |
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 cyclotron is used to accelerate
- Two discs of moment of inertia l₁ and I₂ and angular speeds
- Copper and carbon wires are connected in series and the combined resistor is kept
- A concave mirror ia placed over a beaker containing water of
- The binding energy of deutron is 2.2 MeV and that of ₂⁴He is 28 MeV.
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A cyclotron is used to accelerate
- Two discs of moment of inertia l₁ and I₂ and angular speeds
- Copper and carbon wires are connected in series and the combined resistor is kept
- A concave mirror ia placed over a beaker containing water of
- The binding energy of deutron is 2.2 MeV and that of ₂⁴He is 28 MeV.
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