| ⇦ |
| ⇨ |
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 number of water molecules in gypsum and in plaster of paris are respectively
- When two atoms of chlorine combine to form one molecule of chlorine gas
- The weight of iron which will be converted into its oxide(Fe₃O₄) by the action of 18g
- The alcohol manufactured from water gas is
- The boiling point of heavy water is
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The number of water molecules in gypsum and in plaster of paris are respectively
- When two atoms of chlorine combine to form one molecule of chlorine gas
- The weight of iron which will be converted into its oxide(Fe₃O₄) by the action of 18g
- The alcohol manufactured from water gas is
- The boiling point of heavy water is
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