laws

Question

Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is

Moderate

Solution

According to an ideal gas equation, the molecular weight of an ideal gas is

$M=\frac{\rho RT}{P}\left(\text{as}P=\frac{\rho RT}{M}\right)$

where P, T and $\rho $ are the pressure, temperature and density of the gas respectively and R is the universal gas constant.

$\therefore $ The molecular weight of A is

${M}_{A}=\frac{{\rho}_{A}R{T}_{A}}{{P}_{A}}$

and that of B is

${M}_{B}=\frac{{\rho}_{B}R{T}_{B}}{{P}_{B}}$

$\text{Hence, their corresponding ratio is}$

$\frac{{M}_{A}}{{M}_{B}}=\left(\frac{{\rho}_{A}}{{\rho}_{B}}\right)\left(\frac{{T}_{A}}{{T}_{B}}\right)\left(\frac{{P}_{B}}{{P}_{A}}\right)$

$\text{Here,}\frac{{\rho}_{A}}{{\rho}_{B}}=1.5=\frac{3}{2},\frac{{T}_{A}}{{T}_{B}}=1\text{and}\frac{{P}_{A}}{{P}_{B}}=2$

$\therefore \frac{{M}_{A}}{{M}_{B}}=\left(\frac{3}{2}\right)\left(1\right)\left(\frac{1}{2}\right)=\frac{3}{4}$

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