Question

the ratio of
partial pressure of CH4 and O2 at 500 torr and 100 torr pressure in 10dm3 vessel is/
?

Answer 1

Answer:

Firstly to find the pressure, begin by finding the Mr of methane and oxygen

Mr is the relative molecular mass

Mr of methane:

The formula is - CH4

Mr = 12×1 + 1×4

= 12 + 4

= 16

Mr of oxygen:

The formula is O2 ( 2 is in the subscript)

Mr = 16×2

= 32

So the ratio of the masses of methane : oxygen

= 16 : 32

= 1 : 2

To calculate the force, we will need to multiply the mass with the gravitational field strength that is approximately 9.8N/kg but here both the molecules will experience the same field strength so we can ignore it.

So the ratio of mass is

CH4 : O2

= 1 : 2

Now as given in the question that the number of molecules are in the ratio

methane : oxygen

= 2 : 1

Then multiply the mass and number of molecules ratio

For methane:

= mass × molecules

= 1 × 2

= 2

For oxygen:

= mass × molecules

= 2 × 1

= 2

Thus the pressure is in the ratio =

methane : oxygen

= 2 : 2

= 1 : 1

So as an extension to this thought, the pressure exerted by oxygen and methane will be equal.

Therefore, oxygen exerts 1÷2 or 50% pressure