| ⇦ |
| ⇨ |
The work done during the expansion of a gas from a volume of 4 dm³ to 6 dm³ against a constant external pressure of 3 atm is (1atm = 101.32 J)
Options
(a) -6 J
(b) -608 J
(c) ⁺304 J
(d) -304 J
Correct Answer:
-608 J
Explanation:
Work = -P(ext) x volume change =
-3 x 101.32 x (6 – 4)
= -6 x 101.32 = -607.92 J = -608 J
Related Questions: - The surface tension of which of the following liquid is maximum
- Which of the following oxides is amphoteric in character?
- At 25⁰C the pH value of a solution is 6, then the solution is
- The coordination number of an atom in the simple cubic unit cell in a crystal lattice
- Which of following represents soap
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The surface tension of which of the following liquid is maximum
- Which of the following oxides is amphoteric in character?
- At 25⁰C the pH value of a solution is 6, then the solution is
- The coordination number of an atom in the simple cubic unit cell in a crystal lattice
- Which of following represents soap
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
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