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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 kJ`B. `+99` kJC. `+22.98` kJD. `-22.98` kJ |
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Answer» Correct Answer - B `n=2 moles, V_(1) = 1m^(3) = 10 + 3 dm^(3), V_(2) = 10 dm^(3), pi =100 k Pa` `W=-P_(ex) (V_(2)-V_(1))` `=-100 kPa (10 dm^(3) -1000 dm^(3))` `=-100 kPa (-990 dm^(3))` `+100 xx 990 J` `(therefore kPa xx dm^(3) =J)` `=+99 kJ` |
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