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Question: Pressure of an ideal gas A at \({27^o}C\) is found to be \(2bar\) . When \(2g\) of another ideal gas...

Pressure of an ideal gas A at 27oC{27^o}C is found to be 2bar2bar . When 2g2g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3bar3bar. Find a relationship between their molecular masses.

Explanation

Solution

An ideal gas is a hypothetical gas which follows all the gas laws under ideal conditions of temperature and pressure. In an ideal gas, there is no intermolecular force of attraction among the molecules of the gas and the volume occupied by the molecules is considered negligible with respect to the volume of the container.

Complete step by step answer:
As per the question, let there be two ideal gases A and B.
Let the molecular mass of ideal gas A = MA{M_A}
Let the molecular mass of ideal gas B = MB{M_B}
Given weight of A = wA=1g{w_A} = 1g
Given weight of B = wB=2g{w_B} = 2g
Number of moles of A = wAMA=1MA=nA\dfrac{{{w_A}}}{{{M_A}}} = \dfrac{1}{{{M_A}}} = {n_A}
Number of moles of B = wBMB=2MB=nB\dfrac{{{w_B}}}{{{M_B}}} = \dfrac{2}{{{M_B}}} = {n_B}
Let the pressures of A be = PA=2bar{P_A} = 2bar
Let the pressures of B be = PB{P_B}
The total pressure after B is added to A = 3bar3bar
PA+PB=3bar{P_A} + {P_B} = 3bar
Hence, PB=1bar{P_B} = 1bar
The ideal gas equation for the ideal gases A and B can be written as:
PAV=nART{P_A}V = {n_A}RT ….(i)
PBV=nBRT{P_B}V = {n_B}RT ….(ii)
Dividing equation (i) by equation (ii), we have:
PAPB=nAnB\dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{{n_A}}}{{{n_B}}}
Substituting the value of number of moles, we have:
PAPB=1MA2MB=MB2×MA\dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{\dfrac{1}{{{M_A}}}}}{{\dfrac{2}{{{M_B}}}}} = \dfrac{{{M_B}}}{{2 \times {M_A}}}
Thus, MBMA=2PAPB\dfrac{{{M_B}}}{{{M_A}}} = 2\dfrac{{{P_A}}}{{{P_B}}}
Substituting the values of PA{P_A} and PB{P_B} in the above equation, we have:
MBMA=2×21=4\dfrac{{{M_B}}}{{{M_A}}} = 2 \times \dfrac{2}{1} = 4
Hence, MB=4MA{M_B} = 4{M_A}.

Note: The molar mass of second ideal gas is four times that of the molar mass of the first ideal gas. The number of moles A and B are related to each other in the ratio of 2:12:1 . Thus, the number of molecules of A is present in the container is double the number of molecules of B in the closed container. The volume of the container is constant because of this closed container itself.