Question

A \[0.5d{m^3}\] flask contains gas \[A\] and \[1d{m^3}\] flask contains gas \[B\] at the same temperature. If density of \[A = 3g/d{m^3}\]and that of \[B = 1.5g/d{m^3}\] and the molar mass of \[A = \dfrac{1}{2}of{\text{ }}B\], the ratio of pressure exerted by gases is:
(A) $\dfrac{{{P_A}}}{{{P_B}}} = 2$
(B) $\dfrac{{{P_A}}}{{{P_B}}} = 1$
(C) $\dfrac{{{P_A}}}{{{P_B}}} = 4$
(D) $\dfrac{{{P_A}}}{{{P_B}}} = 3$

Answer 1

Hint: The pressure exerted by a gas depends on volume, temperature and amount of a gas. This is expressed in the form of gas laws of ideal gases which are known as Boyle’s law, Charles’s law and Avogadro’s law.

Complete step by step answer:
According to the kinetic theory of gases, the particles of the gas are in random motion colliding with each other and with the walls of the container. The collisions are elastic in nature and no kinetic energy of the particles is lost. The pressure of a gas is thus explained with the kinetic theory of gases.
Based on the postulates of kinetic theory of gas three laws were proposed relating the pressure, volume, temperature and number of moles (amount of gas) of gas.
Boyle’s law indicates the volume of a gas is inversely proportional to the pressure. Charles’s law indicates the volume of a gas is directly proportional to the temperature. Avogadro’s law indicates the volume of a gas is directly proportional to the moles of gas.
Combining the three laws leads to a general mathematical equation which is known as the ideal gas equation. The equation is
$PV = nRT$ where \[P\] is pressure, \[V\] is volume, \[n\] is moles of gas, \[R\] is gas constant and \[T\] is absolute temperature.
The density (\[d\]) of a gas is equal to the ratio of molar mass (\[M\]) and volume of gas.
Thus \[d = \dfrac{M}{V}\] or \[V = \dfrac{M}{d}\]

Thus the ideal gas equation is \[PM = dRT\],

Inserting the values for gas \[A\] and\[B\] ,

${P_A} = \dfrac{{{d_A}RT}}{{{M_A}}} = \dfrac{{3RT}}{{{M_A}}}$

$\Rightarrow {P_B} = \dfrac{{{d_B}RT}}{{{M_B}}} = \dfrac{{1.5RT}}{{{M_B}}}$

Therefore, $\dfrac{{PA}}{{PB}} = \dfrac{{\dfrac{{3RT}}{{{M_A}}}}}{{\dfrac{{1.5RT}}{{{M_B}}}}} = \dfrac{{2{M_B}}}{{{M_A}}}$

Given, \[{M_A} = \dfrac{1}{2}{M_B}\]

Thus \[\dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{2{M_B}}}{{\dfrac{1}{2}{M_B}}} = 4\] .

So, the correct answer is Option C.

Note: The gases which show deviation from ideal gas behaviour are known as real gases. The pressure of the gas is referred to as the force of colliding particles per unit area. Thus the pressure of a gas is directly proportional to the frequency of collisions per unit time and area of the container.