An enclosure of volume V contains a mixture of 8 g of oxygen, 14 g nitrogen and 22 g of carbon dioxide at absolute temperature T. The pressure of the mixture of gases is (R is universal gas constant)
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a
RTV
b
3RT2V
c
5RT4V
d
7RT5V
answer is C.
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Detailed Solution
Mass of oxygen = 8 gMolecular mass of O2=32 gNumber of moles of O2=832=14=0.25Mass of nitrogen = 14 gMolecular mass of nitrogen gas = 28 gNumber of moles of nitrogen gas =1428=0.5 molesMass of carbon dioxide = 22 gMolecular mass of carbon dioxide = 44 gNumber of moles carbon dioxide =2244=0.5 molesTotal number of moles of the mixture is 0.5+0.5+0.25 μmix=1.25We know that PV=μmixRT Pmix=μmixRTV=1.25 RTV=54RTV