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Calculate the work done during compression of 2 mol of an ideal gas from a volume of `1m^(3)` to `10 dm^(3)` 300 K against a pressure of 100 KPa .A. 99 kJB. `-99` kJC. `114.9` kJD. `-114.9` kJ |
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Answer» Correct Answer - A n=2 moles, `V_(1) = 1m^(3) = 10^(3)dm^(3), V_(2) = 10 dm^(3), P_(ex) = 100 kPa` For isothermal Irreversible process `W = -Pex (V_(2)-V_(1))` `=-100 kPa (10 dm^(3) -1000 dm^(3))` `=-100 (-900) kPa xx dm^(3) = 99000 J` = 99 kJ |
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