Question

Two separate bulbs contain ideal gases A and B. The density of A is twice as that of gas B. The molecular mass of gas A is half as that of B. If two gases are at same temperature, the ratio of the pressure of A to that of B is:
A. $2$
B. $\dfrac{1}{2}$
C. $4$
D. $\dfrac{1}{4}$

Answer 1

We are given two separate bulbs which have ideal gases namely A and B. We have to find the ratio of pressure exerted by the gas when the density of one gas is twice of the other and the molecular mass of one gas is half of the other. We will find the relation between density, molecular mass and pressure of gas by using the gas equation.
Formula Used:
${\text{ PV = nRT}}$

Complete answer:
Since we are given two bulbs which contain ideal gases, therefore we will use the ideal gas equation. This can be represented as,
${\text{ PV = nRT}}$
Where, P is the pressure of gas, V is the volume of ideal gas, n is the number of moles, R is gas constant and T is the temperature of gas. We also know that number moles can be represented as,
$n{\text{ = }}\dfrac{{{\text{given mass}}}}{{{\text{ molar mass}}}}$
Molar mass is represented by M and given mass is represented by m. Also density can be represented as,
${\text{density = }}\dfrac{{{\text{given mass}}}}{{{\text{ volume}}}}$
Density of gas is represented as D and volume is represented as V. Therefore the ideal gas can be deduced as,
${\text{ PV = nRT}}$
${\text{ PV = }}\dfrac{{\text{m}}}{{\text{M}}}{\text{RT}}$
On interchanging the position of M and V we get,
${\text{ PM = }}\dfrac{{\text{m}}}{{\text{V}}}{\text{RT}}$
${\text{ PM = DRT}}$ ________$(i)$
Since R is gas constant and temperature is also the same for both bulbs. Therefore we can ignore them and the relation comes to be,
${\text{ P = }}\dfrac{{\text{D}}}{{\text{M}}}$
, where D is the density of gas and M is the molecular mass of the gas. Let us consider the density of gas A be ${{\text{D}}_A}$ and molecular mass of gas be${{\text{M}}_A}$. Similarly for gas B, the density of gas B be ${{\text{D}}_B}$and the molecular mass of gas be ${{\text{M}}_B}$ . Hence the pressure of gas would be ${P_A}$ and${P_B}$.
According to the question we have,
${D_A}{\text{ = 2}}{{\text{D}}_B}$
${M_A}{\text{ = }}\dfrac{1}{2}{M_B}$
Thus using these two relation we can get the relation between pressures of both gases as,
$\dfrac{{{{\text{P}}_A}}}{{{{\text{P}}_B}}}{\text{ = }}\dfrac{{\dfrac{{{{\text{D}}_A}}}{{{{\text{M}}_A}}}}}{{\dfrac{{{{\text{D}}_B}}}{{{{\text{M}}_B}}}}}$
On putting values we get the result as,
$\dfrac{{{{\text{P}}_A}}}{{{{\text{P}}_B}}}{\text{ = }}\dfrac{{2{{\text{D}}_B}{\text{ }} \times {\text{ 2}}{{\text{M}}_B}}}{{{{\text{D}}_B}{\text{ }} \times {\text{ }}{{\text{M}}_B}}}$
$\dfrac{{{{\text{P}}_A}}}{{{{\text{P}}_B}}}{\text{ = }}\dfrac{4}{1}$
Hence the ratio of pressure of gas A and gas B comes to be $4:1$. Hence the correct option is C.

Note:
Since the temperature of both the gases are constant that’s why we can ignore it. The value of gas constant is fixed units. Its value changes only when the standard of units changes. Therefore both these terms are ignored. We use the ideal gas equation because both gases are given ideally.