6 g of O2 gas at STP is expanded against atmosphere so that the volume is doubled. Thus, the work done is:
A
22.4 L atm
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B
- 4.2 L atm
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C
5.6 L atm
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D
11.2 L atm
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Solution
The correct option is B - 4.2 L atm Expansion work, W=−pext△V=−Pext(Vfinal−Vinitial)
At STP we have: V=22.4L pext=1atm Moles of O2=632=0.1875mol Vintial=22.4×0.1875=4.2L Vinitial=4.2L Vfinal=2×4.2=8.4L Thus, △V=4.2L So, W=−(1atm×4.2L)=−4.2Latm