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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 of the following statement about the member of any one homologous series
- The maximum valency of an element with atomic number 7 is
- The molecular of CO₂ has angle 180°.It can be explained on the basis of
- The bond dissociation energies of H₂, Cl₂ and HCl are 104, 58, and 103
- Which will show geometrical isomerism
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following statement about the member of any one homologous series
- The maximum valency of an element with atomic number 7 is
- The molecular of CO₂ has angle 180°.It can be explained on the basis of
- The bond dissociation energies of H₂, Cl₂ and HCl are 104, 58, and 103
- Which will show geometrical isomerism
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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