JEE Challenger
More from Kinetic Theory
PhyEasyAdv 2013 • Paper 1

Ratio of Densities of Two Ideal Gases Given Mass and Pressure Ratios

Two non-reactive monoatomic ideal gases possess atomic masses in the ratio 2:32 : 3. If they are confined inside a vessel maintained at a constant temperature and have partial pressures in the ratio 4:34 : 3, the ratio of their densities is:

Options

A

1:41 : 4

B

1:21 : 2

C

6:96 : 9

D

8:98 : 9

Correct

Topics & Concepts

Kinetic TheoryKinetic Theory of Gases

Step-by-Step Solution

To find the ratio of the densities of the two ideal gases, we use the ideal gas equation in terms of density:

PV=nRT=(mM)RTP V = n R T = \left(\frac{m}{M}\right) R T

Dividing both sides by the volume VV, we get: P=(mV)RTM=ρRTMP = \left(\frac{m}{V}\right) \frac{R T}{M} = \frac{\rho R T}{M}

Rearranging for density ρ\rho: ρ=PMRT\rho = \frac{P M}{R T}

Since both gases are enclosed in the same vessel, they share the same temperature TT. Thus, the density is directly proportional to the product of partial pressure and molar (atomic) mass: ρ∝PM\rho \propto P M

Taking the ratio of the densities of the two gases: ρ1ρ2=(P1P2)×(M1M2)\frac{\rho_1}{\rho_2} = \left(\frac{P_1}{P_2}\right) \times \left(\frac{M_1}{M_2}\right)

Given:

  • The ratio of atomic masses: M1M2=23\frac{M_1}{M_2} = \frac{2}{3}
  • The ratio of partial pressures: P1P2=43\frac{P_1}{P_2} = \frac{4}{3}

Substitute these values into the ratio: ρ1ρ2=(43)×(23)=89\frac{\rho_1}{\rho_2} = \left(\frac{4}{3}\right) \times \left(\frac{2}{3}\right) = \frac{8}{9}

Therefore, the ratio of their densities is 8:98 : 9.

Correct Answer: D (8:98 : 9)

PreviousNext