States of Matter

• Volume, V = 1 L
• Temperature, T = 27°C = 300 K
• Mass of CH₄ = 4 g, Molar mass = 16 g/mol → Moles, n₁ = 4/16 = 0.25 mol
• Mass of CO₂ = 4.4 g, Molar mass = 44 g/mol → Moles, n₂ = 4.4/44 = 0.1 mol
• Total moles, n = 0.25 + 0.1 = 0.35 mol
• Gas constant, R = 0.0821 L atm K⁻¹ mol⁻¹

Using the ideal gas equation:
PV = nRT
P × 1 = (0.35) × (0.0821) × (300)
P = 8.62 atm

$\frac{P_{N_2}}{P_{O_2}}=\frac{X_{N_2}.P_T}{X_{O_2}.P_T}=\frac{n{N}_2}{n_{O_2}}$$=\frac{70/28}{27/32}=2.96$$\frac{P_{O_2}}{P_{Ar}}=\frac{n_{O_2}}{n_{Ar}}=\frac{27/32}{3/40}=11.25$

$\frac{P_{N_2}}{P_{O_2}}=\frac{X_{N_2}.P_T}{X_{O_2}.P_T}=\frac{n{N}_2}{n_{O_2}}$$=\frac{70/28}{27/32}=2.96$$\frac{P_{O_2}}{P_{Ar}}=\frac{n_{O_2}}{n_{Ar}}=\frac{27/32}{3/40}=11.25$

• Volume, V = 1 L
• Temperature, T = 27°C = 300 K
• Mass of CH₄ = 4 g, Molar mass = 16 g/mol → Moles, n₁ = 4/16 = 0.25 mol
• Mass of CO₂ = 4.4 g, Molar mass = 44 g/mol → Moles, n₂ = 4.4/44 = 0.1 mol
• Total moles, n = 0.25 + 0.1 = 0.35 mol
• Gas constant, R = 0.0821 L atm K⁻¹ mol⁻¹

Using the ideal gas equation:
PV = nRT
P × 1 = (0.35) × (0.0821) × (300)
P = 8.62 atm

Van der Waals equation:

(P + a/Vₘ²)(Vₘ − b) = RT

Multiplying,

P(Vₘ − b) + (a/Vₘ²)(Vₘ − b) = RT

Multiplying entire equation by Vₘ²:

P Vₘ²(Vₘ − b) + a(Vₘ − b) = RT Vₘ²

P Vₘ³ − P b Vₘ² + aVₘ − ab = RT Vₘ²

P Vₘ³ − (Pb + RT)Vₘ² + aVₘ − ab = 0

So coefficients:

Vₘ³ → P
Vₘ² → −(Pb + RT)
Vₘ → a

Required ratio:

(coefficient of Vₘ²) / (coefficient of Vₘ)

= −(Pb + RT) / a

Pb = 300 × 0.060 = 18
RT = 0.082 × 300 = 24.6

Pb + RT = 42.6

Ratio:

= −42.6 / 6
= −7.1

Potential energy dependence:

Charge–charge interaction ∝ 1/r
Charge–dipole interaction ∝ 1/r²

Thus, charge–dipole falls faster than charge–charge. So statement A is incorrect.

Dipole–dipole (rotating molecules) average interaction ∝ 1/r⁶, not 1/r³. So statement B is incorrect.

Dipole-induced dipole interaction does depend on temperature. so statement C is incorrect.

Nonpolar molecules attract each other due to London dispersion forces, so statement D is correct.

At constant V and T, P ∝ n.
Initial: P1 = 2 atm, moles of X = nX.
After adding Y: P2 = 6 atm, total moles = nX + nY.
(nX + nY)/nX = P2/P1 = 6/2 = 3 ⇒ nY = 2nX.

Given mX = 10 g, mY = 80 g.
Mx = 10/nX, My = 80/nY = 80/(2nX) = 40/nX ⇒ My = 4Mx.

u(rms) ∝ 1/√M, so uX/uY = √(My/Mx) = √4 = 2.
Ratio uX : uY = 2 : 1.

At same temperature and volume: Pressure ∝ number of moles

So, P ∝ n

For gas X, let molar mass of X = Mₓ

nₓ = 10 / Mₓ

Given: Pₓ = 2 atm

After adding gas Y, total pressure = 6 atm

So pressure due to Y:

Pᵧ = 6 − 2 = 4 atm

For gas Y, let molar mass of Y = Mᵧ

nᵧ = 80 / Mᵧ

nₓ / nᵧ = Pₓ / Pᵧ

(10 / Mₓ) / (80 / Mᵧ) = 2 / 4

⇒ (10 Mᵧ) / (80 Mₓ) = 1 / 2

⇒ Mᵧ / (8 Mₓ) = 1 / 2

⇒ Mᵧ = 4 Mₓ

RMS velocity relation,

u ∝ 1 / √M

So, uₓ / uᵧ = √(Mᵧ / Mₓ)

= √(4 Mₓ / Mₓ)
= √4
= 2

Final Answer: 2 : 1