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The work done during the expansion of a gas from a volume of `4dm^(3)` to `6dm^(3)` againts a constant external presuure of `3atm` is `(1atm-L=101.32J)`A. `-6J`B. `-608J`C. `+304J`D. `-304J` |
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Answer» Correct Answer - B Work done `(W)=-P_(ext.)(V_(2))-V_(1))` `=-3xx6(6-4)=-6L-atm` `=-6xx101.32J ( :. 1L atm =101.32J)` `=-607.92~~608J.` |
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