| ⇦ | 
| ⇨ |
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: - Perfectly black body radiates the energy 18 J/s at 300K. Another ordinary body of e=0.8
- A moving coil galvanometer of resistance 100 Ω is converted to ammeter by a resistance
- A focal length of a lens is 10 cm. What is power of a lens in dioptre?
- The horizontal range of a projectile is 4√3 times the maximum height achieved
- If a given volume of water in a 220V heater is boiled in 5 min, then same volume
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Perfectly black body radiates the energy 18 J/s at 300K. Another ordinary body of e=0.8
- A moving coil galvanometer of resistance 100 Ω is converted to ammeter by a resistance
- A focal length of a lens is 10 cm. What is power of a lens in dioptre?
- The horizontal range of a projectile is 4√3 times the maximum height achieved
- If a given volume of water in a 220V heater is boiled in 5 min, then same volume
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