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At 27^(@)C, a cylinder of volume 10 L co...

At `27^(@)C`, a cylinder of volume 10 L contains a gas mixture consisting of 0.4 He, 1.6g `O_(2)&1.4g` `H_(2)`. Determine the total pressure of the mixture and the partial pressure of He in the mixture.

Text Solution

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Total number of moles of He, `O_(2) and N_(2)` gas in the mixture `(n)=(0.4)/(4)+(1.6)/(32)+(1.4)/(28)=0.1+0.05+0.05=0.2` mol
[molar masses of He, `O_(2) and N_(2)` are 4, 32 and `28g*mol^(-1)` respectively]
Given, V=10L, T=(273+27K)=300K
we know PV=nRT or, `P=(nRT)/(V)`
`thereforeP=(0.2molxx0.0821L*atm*mol^(-1)*K^(-1)xx300K)/(10L)=0.4926atm`
`therefore` Total pressure of the gas mixture =0.4926 atm
Number of moles of He gas=0.1
and its mole fraction`=(0.1)/(0.2)=0.5`
`therefore` Partial pressure of He gas in mixture=mole-fraction of He in the mixture`xx`total pressure of the mixture`=0.5xx0.4926=0.2463` atm.
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