InterviewSolution
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InterviewSolution
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(`a`) One mole of nitrogen gas at `0.8 atm` takes `38 s` to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at `1.6 atm` takes `57 s` to diffuse through the same hole. Calculate the molecular formula to the compound. (`b`) The pressure exerted by `12 g` of an ideal gas at temperature `t^(@)C` in a vessel of volume `V litre` is `1 atm`. When the temperature is increased by `10^(@)C` at the same volume, the pressure increases by `10%`. Calculate the temperature `t` and volume `V`. (Molecular weight of the gas is `120`.) |
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Answer» (`a`) The rate of diffusion depends on the following facotrs `r prop p` `r prop sqrt(1//M)` Taking these together, we get `(r_(2))/(r_(1))=(p_(2))/(p_(1))((M_(1))/(M_(2)))^(1//2)` Since `r prop 1//t`, we can write `(t_(1))/(t_(2))=(p_(2))/(p_(1))((M_(1))/(M_(2)))^(1//2)` or `M_(2)=((p_(2))/(p_(1))xx(t_(2))/(t_(1)))^(2)M_(1)` Identifying the subscript `1` with nitrogen and `2` with unknown gas, we get `M_(2)=((1.6)/(0.8)xx(57)/(38))^(2)(28 g mol^(-1))=252 g mol^(-1)` Let the molecular formula of the unknown compound be `XeF_(n)`. We will have `M_(Xe)+nM_(F)=252 g mol^(-1)` i.e., `[131+n(19)] g mol^(-1)=252 g mol^(-1)` or `n=(252-131)/(19)=6.36=6` Hence, the molecular formula of the gas is `XeF_(6)`. (`b`) From the expression `pV=nRT`, we can write `(P_(1))/(P_(2))=(T_(1))/(T_(2))` (since `V` and `n` are constants) For the given data, we get `(1 atm)/(1.1 atm)=((273+t//^(@)C)K)/((273+t//^(@)C+10)K)` or `283+t//^(@)C=1.1(273-t//^(@)C)` or `t//^(@)C=(283-1.1xx273)/(0.1)=-173` or `t=-173^(@)C` For the given system `n=(12 g)/(120 g mol^(-1))=0.1 mol` `T=(273-173 K)=100 K` Hence, `V=(nRT)/(P)` `((0.1 mol)(0.082 L atm K^(-1)mol^(-1))(100 K))/(1 atm)` `=0.82 atm` |
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