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5.7     What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9\; dm^{3} flask at 27 ^{o}C ?

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Given a mixture of 3.2g of methane and 4.4g of carbon dioxide contained a 9dm^3 flask at 27 ^{o}C.

So, the pressure exerted by the mixture will be

P =\frac{n}{V}RT = \frac{m}{M}\frac{RT}{V}

The pressure exerted by the Methane gas,P_{CH_{4}} =(\frac{3.2mol}{16}) \frac{0.0821 dm^3 atm K^{-1}mol^{-1}\times300K}{9 dm^3} = 0.55\ atm

The pressure exerted by the Carbon dioxide gas,

P_{CO_{2}} =(\frac{4.4mol}{44}) \frac{0.0821 dm^3 atm K^{-1}mol^{-1}\times300K}{9 dm^3} = 0.27\ atm

So, the total pressure exerted = 0.55+0.27 = 0.82\ atm

And in terms of SI units, we get,

R = 8.314p\ m^3\ K^{-1}mol^{-1},\ V= 9\times 10^{-3}\ m^3

P = 5.543\times10^4 Pa +2.771\times 10^4Pa = 8.314\times10^4 Pa.

Posted by

Divya Prakash Singh

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