Analyzing the Setup
Imagine a closed vessel kept at a constant temperature. Inside, we have a mixture of two non-reactive monoatomic ideal gases. Let's call them Gas 1 and Gas 2. They share the same volume and temperature, but they have different atomic masses and exert different partial pressures.
Our goal is to find the ratio of their densities. To do this, we need a mathematical bridge that connects pressure, density, temperature, and molar mass.
The Master Equation
We start with the fundamental ideal gas equation:
By replacing the number of moles n with the given mass m divided by the molar mass M, we get:
Recognizing that density ρ is simply mass over volume (ρ=Vm), we can rearrange the equation to isolate pressure:
Finally, solving for density ρ, we arrive at the density form of the ideal gas law:
Setting Up the Ratio
Since both gases are in the same vessel, they are at the exact same temperature T. The universal gas constant R is also identical for both. So, if we take the ratio of their densities, ρ1 over ρ2, we can write it as:
ρ2ρ1=(RTp2M2)(RTp1M1)
Notice how the RT terms are identical in both the numerator and the denominator. They perfectly cancel each other out. This simplifies our ratio to just the product of the pressure ratio and the molar mass ratio:
ρ2ρ1=(p2p1)(M2M1)
Final Calculation
Now, let's bring in the values given in the question. The ratio of their partial pressures, p2p1, is 34. And the ratio of their atomic masses, M2M1, is 32. Let's substitute these fractions into our simplified equation:
Multiplying these fractions is straightforward. The numerators multiply to give 4×2=8, and the denominators multiply to give 3×3=9.
So, the ratio of their densities is 8:9. This simple proportional relationship, ρ∝pM, holds beautifully for non-reactive ideal gases at a constant temperature.