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Three moles of an ideal gas are expanded isothermally from a volume of 300 `cm^(3)` to 2.5 L at 300 K against a pressure of 1.9 atm: The work done in joules isA. `-423.56` JB. `+423.56` JC. `-4.18 J`D. `+4.8 J` |
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Answer» Correct Answer - A Given, `V_(1)=300cm^(3)=3xx10^(-4)m^(3)` `V_(2)=2.5 L = 25xx10^(-4)m^(3)` T=300 K `p=1.9 atm =1.9 xx1.01325xx10^(5)N//m^(2)` Now, work done during change in volume against constant pressure is `W=-p(V_(2)-V_(1))` `=-1.9xx1.01325xx10^(5)N//m^(2)` `(25xx10^(-4)-3xx10^(-4))m^(3)` `=-1.925xx10^(5)(22xx10^(-4))Nm` `rArr-423.56 Nm=-423.56 J` `(because1Nm=1J)` |
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