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.
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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 5: 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....
Comments
Why it can't be just p directly proportional to molecular weight by which mw1=2mw2
Greatest explanation
Thank you. Really helpful ð
R is here universal gas constant
R is here universal gas constant
Why does the volume is constant here ... it is not given in question ..????
W means n
R means 0.281 atm
Anita are u know the formula of mAss\molarmass =n and here n is represented by W ðð
What is meaning of wA/mA+wB/mB in this question please answer