States of Matter

5.15. Molar mass of O₂ = 32 g/mol

It means, 8 g of O₂ has 8/32 mol = 0.25 mol

Molar mass of H₂ = 2 g/mol

It means, 4 g of H₂ has 4/2 mol = 2 mol

Therefore, total number of moles, n = 2 + 0.25 = 2.25 mol

Given, V = 1dm³, T = 27°C = 300 K, R = 0.083 bar dm³ K^-1 mol^-1

Applying PV = nRT,

P = nRT / V

   = (2.25) (0.083 bar dm³ K^-1 mol^-1) (300 K) / (1dm³)

   = 56.025 bar

5.14. Time taken to distribute 10¹⁰ wheat grains = 1s

Time taken to distribute Avogadro number of wheat grains = (1s x 6.022 x 10²³) / 10¹⁰

= 6.022 x 10¹³ s

= (6.022 x 10¹³ / 60 x 60 x 24 x 365) year

= 1.9 x 10⁶ years

5.13. Molecular mass of N₂ = 28g
28 g of N₂ has No. of molecules = 6.022 x 10²³ 

1.4 g of N₂ has No. of molecules = (6.022 x 10²³ x 1.4 g)/28 g= 3.011 x 10²² molecules.
Atomic No. of Nitrogen (N) = 7
1 molecule of N₂ has electrons = 7 x 2 = 14
3.011 x 10²² molecules of N₂ have electrons= 14 x 3.011 x 10²²= 4.215 x10²³ electrons.

5.12. Given,

P= 3.32 bar

V= 5 dm³ 

n= 4 mol

R= 0.083 bar dm³ K^-1 mol^-1

PV = nRT

Or T = PV / nR = 3.32 x 5 dm³ / (4.0 mol x 0.083 bar dm³ K^-1 mol^-1)

         = 50 K

5.8. Calculation of partial pressure of H₂ in 1L vessel:

P₁= 0.8 bar, P₂=? V₁= 0.5 L, V₂ = 1.0 L
As temperature remains constant, P₁V₁ = P₂V₂
=> (0.8 bar) (0.5 L) = P₂  (1.0 L)

=>P₂ = 0.40 bar, i.e., P_H2 = 0.40 bar
Calculation of partial pressure of O₂ in 1 L vessel
P₁V₁ = P₂V₂
(0.7 bar) (2.0 L) = P₂  (1L)

=>P₂ = 1.4 bar

=>Po₂= 1.4 bar
Total pressure =P_H2 + P_Q2 = 0.4 bar + 1.4 bar = 1.8 bar

5.5. Suppose molecular masses of A and B are M_A and M_B respectively. Then their number of moles will be

n_A= 1/M_A                                n_B = 2/M_B

Given: P_A = 2 bar        and P_A + P_B = 3 bar

          => P_B = 1bar

Since, PV = nRT

P_A= n_ART and P_BV= n_BRT

Therefore, (P_A / P_B) = (n_A / n_B)= (M_B / 2M_A)

=> M_B / M_A = 2 x P_A / P_B = 2 x 2 /1 = 4

        => M_B = 4 M_A