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# Q5.5. Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of anotherideal gas B is introduced in the same flask at same temperature the pressurebecomes 3 bar. Find a relationship between their molecular masses.

## 3 Answers

9 years ago

let molecular masses be m1 and m2,

PV=nRT,P=(RT/V)(n1 + n2)

2 bar=(RT/V)*n1=(RT/V)*(1/m1)

3 bar=(RT/V)((1/m1)+(2/m2))

9 years ago

The Pressure of A + Pressure of B = 3 Bar (Given :Pressure when both gases are present in the flask under same conditions is 3bar)

Using Daltons law of partial Pressures, Pressure of B = 3-2=1bar

According to an ideal gas equation, For gas A

P1V=n1RT ==> P1V=(m1/M1)RT

Similarly for the gas B

P2V=n2RT ==> P2V=(m2/M2)RT

Diving both the Equ. We get P1/P2=m1M2/m2M1 ==>P1m2/P2m1=M2/M1

M2/M1= P1m2/P2m1

So the ratio of their Molecular masses(M2/M1)=(2*2)/(1*1)=(4:1)---(Ans) Kushagra Madhukar
askIITians Faculty 629 Points
8 months ago
Dear student,
Please find the solution to your problem.

let molecular masses be m1 and m2,
PV = nRT, P = (RT/V)(n1 + n2)
2 bar = (RT/V)*n1 = (RT/V)*(1/m1)  ---(1)
3 bar = (RT/V)((1/m1)+(2/m2))    ---(2)
Dividing equation 2 by 1
Hence,
3/2 = 1 + 2m1/m2
or, m1/m2 = ¼
or, m1 : m2 = 1 : 4

Thanks and regards,
Kushagra

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