+1 vote
in Chemistry by (29.8k points)

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. 

1 Answer

+1 vote
by (1.1k points)
selected by
Best answer

Mass of gas A , WA = 1g

Mass of gas B,  W= 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/M& 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


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

4MA = MB   

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.