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^(@)C ?

What will be the pressure exerted by a mixture of 3.2 g of methane and

1.	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^(@)C ?
Answer» Solution :Calculation of Partial Pressure : Methane `(CH_(4))=3.2` gm Molecular mass of methane = 16 g `mol^(-1)` MOLE of `CH_(4)=(3.2)/(16)=0.2 = n` `V=9 dm^(3)=9xx10^(-3)m^(-3)` `T=(27+273)=300K` `R=8.314 p_(a)m^(3)K^(-1)mol^(-1)` If partial pressure of methane is `p_(SH_(4))` then, `p_(CH_(4))=(n)/(V)RT` `= ((3.2 mol)(8.314 p_(a)m^(3)K^(-1)mol^(-1))(300 K))/(16xx9xx10^(-3)m^(3))` `= 55.43xx10^(3)p_(a)=5.543xx10^(4)p_(a)` Calculation of Partial Pressure of `CO_(2)` : Weight of `CO_(2)(m)=4.4 g` Molecular mass of `CO_(2)=12+2(16)=44g mol^(-1)` Mole of `CO_(2)(n)=(4.4g)/(44 g mol^(-1))=0.1` mol If Partial Pressure of `CO_(2)` is `p_(CO_(2))` then, `p_(CO_(2))=(n)/(V)RT` `= ((0.1 mol)(8.314 p_(a)m^(3)K^(-1)mol^(-1))(300 K))/(9xx10^(-3)m^(3))` `= 27.713xx10^(3)p_(a)=2.7713xx10^(4)p_(a)` `p_("TOTAL")=(p_(CO_(2))+p_(CH_(4)))` `= (2.7713xx10^(4)+5.543xx10^(4))p_(a)` `= 8.314xx10^(4)p_(a)` OR `p_("total")=(p_(CH_(4))+p_(CO_(2)))` `= ((0.2+0.1)(8.314 p_(a)m^(3)K^(-1)mol^(-1))(300K))/(9xx10^(-3)m^(3))` `= 8.314xx10^(4)p_(a)`