Question

A 10 L flask contains 2g of CH4,3g of hydrogen and 4g of nitrogen at 20 ˚ C. a. What is the pressure, in atm inside the flask? b. What is the partial pressure of each component of the mixture of gases?

Answer 1

a) `4.222atm`.
b) Methane: `0.301atm`, hydrogen: `3.578atm` and nitrogen: `0.344atm`.
I profusely apologize for the long answer, but I knew no shorter way. I assure you, half of it is empty spaces and calculations to make it easier.

Explanation:

a)
We have `2g` of `CH_4`, `3g` of `H_2` and `4g` of `N_2`. Since the gases take up the entire volume, the volume is `10L` and the temperature `20^@C`.

Use the Ideal Gas Law (`PV=nRT`):
Known values:
`V=10L`
`R=0.082057338 L atm K^-1 mol^-1`
`T=20+273=293K`

We have to find `P`, but we're now missing the value of `n`, the moles. Let's calculate:
Molar Mass:

• Methane - `16.043g`
• Hydrogen - `2.016g`
• Nitrogen - `28.014g`

Now using the formula for moles (`"given mass"/"molecular mass"`), calculate:

The moles of methane: `2/16.043=0.125mol`
The moles of hydrogen: `3/2.016=1.488mol`
The moles of nitrogen: `4/28.014=0.143mol`

Add them up to find the total moles of gas: `0.125+1.488+0.143=1.756mol`

So we have `10L` of gas, or `1.756mol`, at `293K`.

The formula for pressure using the ideal gas law is:
`(nRT)/V`

`(1.756*0.082057338*293)/10`

`42.22/10`

`4.222`

So the pressure is (to three significant figures) `4.222atm`.

b)
We have to input the different amounts of moles into the formula. Quite simply, the partial pressure of a particular gas is the pressure if that gas alone occupied the flask.
In this question, we can use our earlier formula: `(nRT)/V`, but now changing the value of `n` based on the gas.
We can confirm this answer using the formula `P_"total"=P_1+P_2+P_3......+P_n`. In this situation, only three partial pressures are there. This is the Law of Partial Pressures.

Methane:
`(nRT)/V`

There are `0.125mol` of methane, so:
`(0.125*0.082057338*293)/10`

`3.005/10`

`0.301atm`

The methane exerts a pressure of `0.301atm`.

Hydrogen:
`(nRT)/V`

There are `1.488mol` of hydrogen, so:
`(1.488*0.082057338*293)/10`

`35.778/10`

`3.578atm`

The hydrogen exerts a pressure of `3.578atm`.

Nitrogen:
`(nRT)/V`

There are `0.143mol` of hydrogen, so:
`(0.143*0.082057338*293)/10`

`3.438/10`

`0.344atm`

The nitrogen exerts a pressure of `0.344atm`.

To confirm, add up the individual pressures:
`0.301+3.578+0.344=4.223atm`, equal to the total pressure.

You may be confused. Why `4.223atm`? This is because of my rounding, but if you use an extended calculation without rounding, this will be the answer.