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Q8. Graham's Diffusion (Effusion) Law.
We know a mole of ideal gas occupies 22.4 L at STP.
1.12 L of H2 at STP contains 1.12/22.4 = 0.05 moles (0.10 gms)
1.12 L of D2 at STP contains 1.12/22.4 = 0.05 moles of D2 (0.20 gms)
Molar mass of H2 = 2. Molar mass of D2 = 4
Gases H2 and D2 diffuse (effuse) into 2nd bulb, at rates r1 and r2 respectively.
r1 / r2 = √(4/2) => r1 = √2 * r2
=> number of moles of H2 in bulb2 will be √2 times number of moles of D2.
Number of moles of H2 remaining in bulb1 = 0.05 gm / (2gm/mol) = 0.025
So number of moles of H2 that diffused into bulb2 = 0.05 - 0.025 = 0.025
So number of moles of D2 that diffused into bulb2 = 0.025/√2 =
Weight of D2 in bulb2 = 0.025/√2 * 4 = 0.05 √2 gms
Total weight of gases in bulb2: 0.025 * 2 + 0.05 √2 = 0.05 *(1+√2) gm
Composition of H2 in bulb2 = 0.05/[ 0.05(1+√2)] * 100
= 100/(1+√2) % = (√2-1)*100 = 41.4% by weight
Composition of D2 in bulb2: √2 /(1+√2) = (2-√2)*100 % by weight
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Q9.
Compressibility Factor = Z
Z = Volume of real gas / Volume of Ideal gas , at same T & P.
Z = P*V/(n R T)
=> Real Volume = V = nRT * Z / P
V1 = Real Volume of N2 of n moles at T1 = 223K, P1 = 800 atm
V2 = Real Volume of N2 of n moles at T2 =373K, P2 = 200 atm.
Z1 = 1.95. Z2 = 1.10.
V2/V1 = [ T2 Z2 P1 ] / [T1 Z1 P2 ]
= [ 373* 1.10 * 800 ] / [223 * 1.95 * 200 ]
= 3.43
Given at T=223 K and P=800 atm , V1 = 1.0 dm³
So Volume of same quantity of N2 at P = 200 atm and 373K
= 3.43 * 1 dm³ = 3.43 dm³
====================
Q10. Graham's diffusion law:
Assume the mixture of H2 and Gas G to be Ideal. Find the number n of moles in it.
P * V = n R T
6 atm * 3 Litre = n * 8.314 J/°K/mol * (273+27)°K
n = 6*1.013*10⁵ * 3 * 10⁻³ / [ 8.314 * 300°K]
= 0.731 moles.
Number of moles of H2 = 0.7 moles.
Hence the number of moles of Gas G = 0.731 - 0.7 = 0.031 moles.
Ratio of Rates of diffusion of H2 & Gas G = 0.7 / 0.031
As per Graham's law: 0.7/0.031 = √ [ M / 2] , M = molar mass of Gas G.
=> M = 1019.77 gms/mole
We know a mole of ideal gas occupies 22.4 L at STP.
1.12 L of H2 at STP contains 1.12/22.4 = 0.05 moles (0.10 gms)
1.12 L of D2 at STP contains 1.12/22.4 = 0.05 moles of D2 (0.20 gms)
Molar mass of H2 = 2. Molar mass of D2 = 4
Gases H2 and D2 diffuse (effuse) into 2nd bulb, at rates r1 and r2 respectively.
r1 / r2 = √(4/2) => r1 = √2 * r2
=> number of moles of H2 in bulb2 will be √2 times number of moles of D2.
Number of moles of H2 remaining in bulb1 = 0.05 gm / (2gm/mol) = 0.025
So number of moles of H2 that diffused into bulb2 = 0.05 - 0.025 = 0.025
So number of moles of D2 that diffused into bulb2 = 0.025/√2 =
Weight of D2 in bulb2 = 0.025/√2 * 4 = 0.05 √2 gms
Total weight of gases in bulb2: 0.025 * 2 + 0.05 √2 = 0.05 *(1+√2) gm
Composition of H2 in bulb2 = 0.05/[ 0.05(1+√2)] * 100
= 100/(1+√2) % = (√2-1)*100 = 41.4% by weight
Composition of D2 in bulb2: √2 /(1+√2) = (2-√2)*100 % by weight
=============
Q9.
Compressibility Factor = Z
Z = Volume of real gas / Volume of Ideal gas , at same T & P.
Z = P*V/(n R T)
=> Real Volume = V = nRT * Z / P
V1 = Real Volume of N2 of n moles at T1 = 223K, P1 = 800 atm
V2 = Real Volume of N2 of n moles at T2 =373K, P2 = 200 atm.
Z1 = 1.95. Z2 = 1.10.
V2/V1 = [ T2 Z2 P1 ] / [T1 Z1 P2 ]
= [ 373* 1.10 * 800 ] / [223 * 1.95 * 200 ]
= 3.43
Given at T=223 K and P=800 atm , V1 = 1.0 dm³
So Volume of same quantity of N2 at P = 200 atm and 373K
= 3.43 * 1 dm³ = 3.43 dm³
====================
Q10. Graham's diffusion law:
Assume the mixture of H2 and Gas G to be Ideal. Find the number n of moles in it.
P * V = n R T
6 atm * 3 Litre = n * 8.314 J/°K/mol * (273+27)°K
n = 6*1.013*10⁵ * 3 * 10⁻³ / [ 8.314 * 300°K]
= 0.731 moles.
Number of moles of H2 = 0.7 moles.
Hence the number of moles of Gas G = 0.731 - 0.7 = 0.031 moles.
Ratio of Rates of diffusion of H2 & Gas G = 0.7 / 0.031
As per Graham's law: 0.7/0.031 = √ [ M / 2] , M = molar mass of Gas G.
=> M = 1019.77 gms/mole
:-)
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