States of Matter

5.25.  (a) Other than attractive forces, molecules also exert repulsive forces on one another. When two molecules are brought into close contact with each other, the repulsion between the electron clouds and that between the nuclei of two molecules comes into play. Magnitude of the repulsion rises very rapidly as the distance separating the molecules decreases. This is the reason that liquids and solids are hard to compress. In these states molecules are already in close contact; therefore, they resist further compression; as that would result in the increase of repulsive interactions.

5.24.  (b) Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.

5.23.  (a) The dipole-dipole interaction between two HCl molecules is stronger than the London forces but is weaker than ion-ion interaction because only partial charges are involved.

5.22. Higher the critical temperature, more easily the gas can be liquefied, i.e., greater are the intermolecular forces of attraction. Hence, CO₂ has stronger intermolecular forces than CH₄.

5.21. At -273°C, volume of the gas becomes equal to zero, i.e., the gas ceases to exist.

5.20. SI unit of pressure, P = Nm^-2

SI unit of volume = m³

Si unit of temperature, T = K

SI unit of number of moles, n = mol

Thus, SI unit of pV²T²/n = (Nm^-2) (m³)² (K) ²mol = Nm⁴K²mol^-1

5.19. As the mixture H₂ and O₂ contains 20% by weight of dihydrogen, therefore,

If H₂ = 20g, then O₂ = 80g

No. of moles of H₂ = 20/2 = 10 moles

No. of moles of O₂ = 80/32 = 2.5 moles

Partial pressure of H₂ = [No. of moles of H₂/ (No. of moles of H₂ + No. of moles of O₂)] x

P_total= [10 (10 + 2.5)] x 1 = 0.8 bar

5.18. Given, P₁ = P₂ and V₁ = V₂

We know that P₁V₁ = P₂V₂

Or,                n₁RT₁ = n₂RT₂

i.e.,                  n₁T₁ = n₂T₂

 Substituting n = w/M, we get

(W₁/M₁) x T₁ = (W₂/M₂) x T₂

(2.9/M₁) x (95 + 273) = (0.184/2) x (17 + 273)

M₁ = (2.9 x 368 x 2) / (0.184 x 290) = 40 g mol^-1

5.17. No. of moles of CO₂ = Given mass of CO₂ / Molar mass

                                           = 8.8g / 44g mol^-1 = 0.2 mol

Pressure of CO₂ = 1 bar

RR = 0.083 bar dm³ K^–1 mol^–1

T = 273 + 31.1 K = 304.1 K

According to ideal gas equation,

PV = nRT

Therefore, V = nRT/P

                      = (0.2 x 0.083 x 304.1) / 1 bar = 5.048 L