| ⇦ | 
| ⇨ |
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: - According to Dalton’s atomic theory the smallest partical of the matter is called
- Oxidation of acetaldehyde with selenium dioxide produces
- The indicator used in the titration of iodine against sodium thiosulphate is
- The coordination number and oxidation state number of Cr in K₃ Cr (C₂O₄)₃
- Which of the following reagents can be used to convert primary amides into pri
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- According to Dalton’s atomic theory the smallest partical of the matter is called
- Oxidation of acetaldehyde with selenium dioxide produces
- The indicator used in the titration of iodine against sodium thiosulphate is
- The coordination number and oxidation state number of Cr in K₃ Cr (C₂O₄)₃
- Which of the following reagents can be used to convert primary amides into pri
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