Question

Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another
ideal gas B is introduced in the same flask at same temperature the pressure
becomes 3 bar. Find a relationship between their molecular masses.
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Answer 1

Answer:

Mass of gas A , WA = 1g

Mass of gas B, WB = 2g

Pressure exerted by the gas A = 2 bar

Total pressure due to both the gases = 3 bar

In this case temperature & volume remain constant

Now if MA & MB are molar masses of the gases A & B respectively,therefore

pA V= WA RT/MA & Ptotal V = (WA/MA + WB/ MB) RT

= 2 X V = 1 X RT/MA & 3 X V = (1/MA + 2/MB)RT

From these two equations, we get

3/2 = (1/MA + 2/MB) / (1 /MA) = (MB + 2MA) / MB

This result in 2MA/ MB = (3/2) -1 = ½

OR MB = 4MA

Thus, a relationship between the molecular masses of A and B is given by

4MA = MB ..

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