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(a) For Dalton's law of partial pressure, derive the expression Pgas = Xgas. Ptotal 

(b) A 2 L flask contains 1.6 g of methane and 0.5 g of hydrogen at 27°C. Calculate the partial pressure of each gas in the mixture and hence, calculate the total pressure.

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(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:

Pgas = Xgas.Ptotal

(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

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