| ⇦ | 
| ⇨ |
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 greenhouse gas is
- Sodium Hydroxide reacts with Zinc to form
- If Ksp for HgSO₄ is 6.4 x 10⁻⁵, then solubility of the salt is
- Which one of the elements with the following outer orbital configuration
- For an endothermic reaction the value of ΔS is positive.The reaction is
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The greenhouse gas is
- Sodium Hydroxide reacts with Zinc to form
- If Ksp for HgSO₄ is 6.4 x 10⁻⁵, then solubility of the salt is
- Which one of the elements with the following outer orbital configuration
- For an endothermic reaction the value of ΔS is positive.The reaction 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