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The density of a gas A is twice that of gas B at the same temperature. The molecular mass of B is thrice that of A. The ratio of the pressures acting on A and B will be ?
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The density of a gas A is twice that of gas B at the same temperature....
We have,
From ideal gas equation,
PV= nRT = (m/M)RT where, m = mass in gm and M = molar mass of the gas.
Further we derive that, P = m/V x RT/M= dRT/M
So, M = d RT/P
Since, in the given case, R and T all are constant so,

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The density of a gas A is twice that of gas B at the same temperature....
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The density of a gas A is twice that of gas B at the same temperature....
Density and Molecular Mass

Density: Density is defined as the mass per unit volume of a substance. It is a measure of how closely packed the particles of a substance are. The density of a gas depends on its mass and volume.

Molecular Mass: Molecular mass is the sum of the atomic masses of all the atoms in a molecule. It is expressed in atomic mass units (amu) or grams per mole (g/mol).

Given Information:
- The density of gas A is twice that of gas B at the same temperature.
- The molecular mass of B is thrice that of A.

Relationship between Density and Molecular Mass:
The density of a gas is directly proportional to its molecular mass. This means that as the molecular mass of a gas increases, its density also increases, assuming the temperature and pressure are constant. This relationship can be explained using the ideal gas law equation:

Pressure and Density:
The pressure exerted by a gas is related to its density according to the ideal gas law equation:

PV = nRT

Where:
- P is the pressure of the gas
- V is the volume of the gas
- n is the number of moles of the gas
- R is the ideal gas constant
- T is the temperature of the gas

Ratio of Pressures:
To find the ratio of the pressures acting on gases A and B, we need to consider their densities and molecular masses.

Let's assume the density of gas A is ρA and the density of gas B is ρB.

Given that the density of A is twice that of B, we can write:
ρA = 2ρB

Given that the molecular mass of B is thrice that of A, we can write:
Molecular mass of A = M
Molecular mass of B = 3M

From the ideal gas law equation, we can rearrange it to solve for pressure:
P = (nRT)/V

Since the number of moles, gas constant, and volume are the same for gases A and B (as they are at the same temperature and we assume the same volume), we can cancel them out.

Therefore, the ratio of the pressures acting on A and B can be calculated as follows:

(PA/PB) = (ρA/Molecular mass of A) / (ρB/Molecular mass of B)

Substituting the given values, we get:
(PA/PB) = (2ρB/M) / (ρB/3M)

Simplifying the equation, we find:
(PA/PB) = 6

Therefore, the ratio of the pressures acting on gases A and B is 6:1.
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