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The work done during the expansion of a gas from a volume of `4dm^(3)` to `6dm^(3)` against a constant external pressure of 3 atm is (1L atm = 101.32 J)A. `+ 304 J`B. `- 304 J`C. `- 6 J`D. `-608 J` |
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Answer» Correct Answer - D `W=-pDeltaV, W=-3xx(6-4)` `W=-6xx101.32(therefore "1 L atm=101.32 J")` W=-608 J |
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