NCERT Class XI Chemistry States of Matter Solutions

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Question : 5
Total: 23
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.
Solution:  
For gas A, PAV =
mA
MA
RT ... (1)
For gas B, PBV =
mB
MB
RT ... (ii)
Dividing equation (1) by equation (2) gives
PA
PB
=
mA
mB
MB
MA

PA+PB = 3 ⇒ PB = 3 - 2 = 1 bar
MA
MB
= (
mA
mB
)
(
PB
PA
)
= (
1g
2g
)
(
1bar
2bar
)
MA
MB
= 1 : 4
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