(a) Suppose, at temperature T, these gases, enclosed in the volume V, exert partial pressure P1, P2 and P3, respectively. Then,
P1 = \(\frac{n_1RT}{V}\) ..............(i)
P2 = \(\frac{n_2RT}{V}\) ............(ii)
P3 = \(\frac{n_3RT}{V}\) ..............(iii)
Where, n1, n2 and n3 = Number of moles of these gases
\(\therefore\) Total pressure Ptotal = P1 + P2 + P3
= n1\(\frac{RT}{V}\)+ n2\(\frac{RT}{V}\)+ n3\(\frac{RT}{V}\)
= (n1+n2+n3)\(\frac{RT}{V}\) ..........(iv)
On dividing P1 by Ptotal, we get,
\(\frac{P_1}{P_{Total}}\)= \(\big(\frac{n_1}{n_1+n_2+n_3}\big)\frac{RTV}{RTV}\)
= \(\frac{n_1}{n_1+n_2+n_3}\) = \(\frac{n_1}{n}\) = x1
Where x1 = mole fraction of first gas
\(\therefore\) P1 = x1Ptotal
Similarly, for other two gases
P2 = X2 Ptotal
P3 = X3 Ptotal
General equation can be written as:
(b) V = 2L, T = 27 + 273 = 300K
Number of moles of CH4
= \(\frac{Mass\,of\,CH_4}{Molar\,mass\,of\,CH_4}\)
= \(\frac{1.6}{16}\) = 0.1 mol
Number of moles of H2
= \(\frac{Mass\,of\,H_2}{Molar\,mass\,of\,H_2}\)
According to ideal gas equation
PV = nRT
Partial pressure of CH4 (PCH4) = \(\frac{nRT}{V}\)
= \(\frac{0.1\,mol\times0.0821\,L\,atm\,K^{-1}mol^{-1}\times300K}{2L}\)
= 1.23 atm
Partial pressure of H2(PH2) = \(\frac{nRT}{V}\)
= \(\frac{0.25\,mol\times0.0821\,L\,atm\,K^{-1}mol^{-1}\times300K}{2L}\)
= 3.079 atm
Total pressure = 1.23 + 3.079 = 4.31 atm