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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: - Which one of the following statements about halogens is correct
- Heavy water is
- At 25⁰ C , the highest osmotic pressure is exhibited by 0.1 M solution of
- 3-Pentanol upon reaction with PBr₃ gives 2 and 3-bromopentane.Such rearrangement
- The most easily hydrolysed molecule under SN1 conditions is
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which one of the following statements about halogens is correct
- Heavy water is
- At 25⁰ C , the highest osmotic pressure is exhibited by 0.1 M solution of
- 3-Pentanol upon reaction with PBr₃ gives 2 and 3-bromopentane.Such rearrangement
- The most easily hydrolysed molecule under SN1 conditions is
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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