| ⇦ |
| ⇨ |
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 thermocouple is made from two metals, antimony and bismuth. If one junction
- The coefficient of performance of a refrigerator is 5. If the temperature inside
- Two bubbles A and B (A>B) are joined through a narrow tube. Then
- Three cells, each of e.m.f 1.5 V and internal resistance 1 ohm are connected in parallel
- To draw the maximum current from a combination of cells, how should be the cells
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A thermocouple is made from two metals, antimony and bismuth. If one junction
- The coefficient of performance of a refrigerator is 5. If the temperature inside
- Two bubbles A and B (A>B) are joined through a narrow tube. Then
- Three cells, each of e.m.f 1.5 V and internal resistance 1 ohm are connected in parallel
- To draw the maximum current from a combination of cells, how should be the cells
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