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Thermodynamics question

2004 · Q98
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Thermodynamics question

2004 · Q98

NEETChemistryThermodynamicsMCQ+4 / −1
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 (1 L atm = 101.32 J)
  1. A
    −-− 6 J
  2. B
    −-− 608 J
  3. C
    + 304 J
  4. D
    −-− 304 J
View written solutionFree

Correct answer: B

Work done during the expansion, W = – pdV

W = –3 atm (6 dm3 – 4 dm3)

= – 3 atm ( 2 dm3 ) (1 dm3 = 1 L)

= – 3 atm × 2 L

= – 6 L atm

As, 1 L atm = 101.32 J

∴\therefore∴ W = – 6 × 101.32 J = – 607.92 J ≈ – 608 J

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