At 15°C, a 1.00 L container contains 6.41 mol $N_{2}$ $N_2$ and 3.82 mol $A r$ $Ar$ . What is the partial pressure of each gas, and what is the total pressure inside the container?

Question

At 15°C, a 1.00 L container contains 6.41 mol $N_{2}$ $N_2$ and 3.82 mol $A r$ $Ar$ . What is the partial pressure of each gas, and what is the total pressure inside the container?

1 Answer

Impact of this question

4549 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-10T20:50:45

Here's what I got.

Explanation:

Here's an alternative approach for which you only have to use the ideal gas law equation once.

According to Dalton's Law of Partial Pressures, the partial pressure of a gas that's part of a gaseous mixture is proportional to the number of moles said gas has in the mixture.

In other words, the partial pressure of a gas that's part of a gaseous mixture depends on the mole fraction of the gas and on the total pressure of the mixture.

You can read more on that in this Socratic answer.

Mathematically, this is written as

$P_{i} = χ_{i} \times P_{total}$ $color(blue)(P_i = chi_i xx P_"total")" "$ , where

$P_{i}$ $P_i$ - the partial pressure of gas $i$ $i$
$χ_{i}$ $chi_i$ - the mole fraction it has in the mixture
$P_{total}$ $P_"total"$ - the total pressure of the mixture

The mole fraction of a gas in a gaseous mixture is defined as the number of moles of that gas divided by the total number of moles present in the mixture.

In your case, the total number of moles will be

$n_{total} = n_{N_{2}} + n_{A r}$ $n_"total" = n_(N_2) + n_(Ar)$

$n_{total} = 6.41 moles + 3.82 moles = 10.23 moles$ $n_"total" = "6.41 moles" + "3.82 moles" = "10.23 moles"$

The mole fractions of the two gases will be

$chi_(N_2) = (6.41 color(red)(cancel(color(black)("moles"))))/(10.23color(red)(cancel(color(black)("moles")))) = 0.6266$

$chi_(Ar) = (3.82 color(red)(cancel(color(black)("moles"))))/(10.23color(red)(cancel(color(black)("moles")))) = 0.3734$

This means that all you need to know now is the total pressure of the mixture $->$ cue the ideal gas law equation!

$color(blue)(PV = nRT implies P_"total" = (n_"total" * RT)/V)$

$P_"total" = (10.23 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 15)color(red)(cancel(color(black)("K"))))/(1.00color(red)(cancel(color(black)("L"))))$

$P_"total" = "242 atm"$

Therefore, you will have

$P_(N_2) = 0.6266 * "242 atm" = "151.6 atm"$

$P_(Ar) = 0.3734 * "242 atm" = "90.4 atm"$

Now, you should round these off to two sig figs, the number of sig figs you have for temperature of the mixture, but I'll leave them as-is

${ (P_(N_2) = color(green)("151.6 atm")), (P_(Ar) = color(green)("90.4 atm")), (P_"total" = color(green)("242 atm")) :}$

Science

Math

Humanities

... and beyond

At 15°C, a 1.00 L container contains 6.41 mol $N_{2}$ $N_2$ and 3.82 mol $A r$ $Ar$ . What is the partial pressure of each gas, and what is the total pressure inside the container?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

At 15°C, a 1.00 L container contains 6.41 mol N_2N2N_2 and 3.82 mol ArArAr. What is the partial pressure of each gas, and what is the total pressure inside the container?

1 Answer

Explanation:

Related questions

Impact of this question

At 15°C, a 1.00 L container contains 6.41 mol $N_{2}$ $N_2$ and 3.82 mol $A r$ $Ar$ . What is the partial pressure of each gas, and what is the total pressure inside the container?