# 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 $27 ^{o}C$ ?

D Divya Prakash Singh

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,

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

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

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