| ⇦ | 
| ⇨ |
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: - The following four wires are made of the same material. Which of these will have
- An electric dipole is placed at an angle 30°with an electric field intensity 2×10⁵N/C.
- A body of mass 0.5 kg travels in a straight line with velocity v=a x³/², where
- A solid which is transparent to visible light and whose conductivity increases
- Copper of fixed volume V is drawn into wire of length l. When this wire is subjected
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The following four wires are made of the same material. Which of these will have
- An electric dipole is placed at an angle 30°with an electric field intensity 2×10⁵N/C.
- A body of mass 0.5 kg travels in a straight line with velocity v=a x³/², where
- A solid which is transparent to visible light and whose conductivity increases
- Copper of fixed volume V is drawn into wire of length l. When this wire is subjected
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