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?

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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?

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Answer 1 · 2017-11-15T11:13:51

a) $4.222 a t m$ $4.222atm$ .
b) Methane: $0.301 a t m$ $0.301atm$ , hydrogen: $3.578 a t m$ $3.578atm$ and nitrogen: $0.344 a t m$ $0.344atm$ .
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 $2 g$ $2g$ of $C H_{4}$ $CH_4$ , $3 g$ $3g$ of $H_{2}$ $H_2$ and $4 g$ $4g$ of $N_{2}$ $N_2$ . Since the gases take up the entire volume, the volume is $10 L$ $10L$ and the temperature $20^{\circ} C$ $20^@C$ .

Use the Ideal Gas Law ( $P V = n R T$ $PV=nRT$ ):
Known values:
$V = 10 L$ $V=10L$
$R = 0.082057338 L a t m K^{- 1} m o l^{- 1}$ $R=0.082057338 L atm K^-1 mol^-1$
$T = 20 + 273 = 293 K$ $T=20+273=293K$

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

Methane - $16.043 g$ $16.043g$
Hydrogen - $2.016 g$ $2.016g$
Nitrogen - $28.014 g$ $28.014g$

Now using the formula for moles ( $\frac{given mass}{molecular mass}$ $"given mass"/"molecular mass"$ ), calculate:

The moles of methane: $\frac{2}{16.043} = 0.125 m o l$ $2/16.043=0.125mol$
The moles of hydrogen: $\frac{3}{2.016} = 1.488 m o l$ $3/2.016=1.488mol$
The moles of nitrogen: $\frac{4}{28.014} = 0.143 m o l$ $4/28.014=0.143mol$

Add them up to find the total moles of gas: $0.125 + 1.488 + 0.143 = 1.756 m o l$ $0.125+1.488+0.143=1.756mol$

So we have $10 L$ $10L$ of gas, or $1.756 m o l$ $1.756mol$ , at $293 K$ $293K$ .

The formula for pressure using the ideal gas law is:
$\frac{n R T}{V}$ $(nRT)/V$

$\frac{1.756 \cdot 0.082057338 \cdot 293}{10}$ $(1.756*0.082057338*293)/10$

$\frac{42.22}{10}$ $42.22/10$

$4.222$ $4.222$

So the pressure is (to three significant figures) $4.222 a t m$ $4.222atm$ .

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: $\frac{n R T}{V}$ $(nRT)/V$ , but now changing the value of $n$ $n$ 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.

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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?

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