A 1.00 L flask contains nitrogen gas. what is the final partial pressure?

A 1.00 L container contains nitrogen gas at 300k and a pressure of 2.3 atm. 0.30 mol of O2(g) is added to the flask and allowed to mix. Then a lid is opened to allow 0.040 moles of molecules to leave.
What is the partial pressure of oxygen in the final mixture?

2 Answers
Jun 16, 2018

Answer:

#"6 atm"#.

Explanation:

To answer this question, we'll need to use two formulas:

  • The ideal gas law, or #pV = nRT#.
  • An expression of Dalton's law of partial pressures, or
    #P_a = x_a xx P_"total"#,
    where #a# is a gas and #x_a# is the mole fraction of that gas.

First, let's use the ideal gas law to find the number of moles of nitrogen in the container! :)

The question tells us that the #N_2# gas is at #"300 K"#, #"2.3 atm"#, and stored in a #"1.00 L"# container.
So, we'll just plug these values into the equation, using #"0.08206 L atm/K mol"# as our gas constant because it fits our units the best:

#pV = nRT#

#n = (pV)/(RT) = ("2.3 atm" xx "1.00 L")/("0.08206 L atm/K mol" xx "300 K")#

#n = "0.09 mol"#

Now, we'll start gathering the values needed for the second equation to solve for the partial pressure of #O_2#.
We need to find the mole fraction of #O_2# first—we know that #"0.30 mol"# of #O_2# was added into the container and then #"0.040 mol"# of gas molecules leave after that.

Let's assume that an equal amount of #N_2# and #O_2# left; that, out of the #"0.040 mol"#, #"0.020 mol"# of #N_2# left and #"0.020 mol"# of #O_2# left.

This would leave us with #"0.09 mol" - "0.020 mol" = "0.07 mol"# of #N_2# gas and #"0.30 mol" - "0.040 mol" = "0.26 mol"# of #O_2# gas.

So, the mole fraction of #O_2# gas would be:

#x_(O_2) = (O_2)/(O_2 + N_2) = ("0.26 mol")/("0.26 mol" + "0.07 mol") = 0.8#

Then, we need to find the total pressure of both #N_2# and #O_2#.
Since the container is still #"1.00 L"#, (assuming it's) still at #"300 K"#, and now has #"0.26 mol" + "0.07 mol" = "0.3 mol"# of gas in it, we can solve for pressure using the ideal gas law:

#pV = nRT#

#p = (nRT)/v = ("0.3 mol" xx "0.08206 L atm/K mol" xx "300 K")/"1.00 L"#

#p = "7 atm"#

Now, we can use an expression of Dalton's law of partial pressures to solve for the partial pressure of #O_2#:

#P_a = x_a xx P_"total"#

#P_(O_2) = x_(O_2) xx P_"total"#

#P_a = 0.8 xx "7 atm" = "6 atm"#

Jun 16, 2018

Answer:

#p_text(O₂) = "6.6 atm"#

Explanation:

Step 1. Calculate the moles of nitrogen

To answer this question, we'll need to use the Ideal Gas Law.

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

We can rearrange this equation to get

#n = (pV)/(RT)#

In this problem,

#pcolor(white)(l) = "2.3 atm"#
#V = "1.00 L"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "300 K"#

#n = (2.3 color(red)(cancel(color(black)("atm"))) × 1.00 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 300 color(red)(cancel(color(black)("K")))) = "0.0934 mol"#

Step 2. Calculate the final moles of oxygen in the container

After adding oxygen, we have

#"0.0934 mol N"_2 + "0.30 mol O"_2 = "0.3934 mol gas"#

After allowing 0.040 mol of gas to leave, we have

#"(0.3934 - 0.040) mol gas = 0.3534 mol gas"#

#" Moles of O"_2 = 0.3534 color(red)(cancel(color(black)("mol gas"))) × "0.30 mol O"_2/(0.3934 color(red)(cancel(color(black)("mol gas")))) = "0.270 mol O"_2#

Step 3. Calculate the partial pressure of the oxygen

We can again use the Ideal Gas Law.

#p = (nRT)/V#

This time,

#n = "0.270 mol"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "300 K"#
#V = "1.00 L"#

# p = (0.270 color(red)(cancel(color(black)("mol"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 300 color(red)(cancel(color(black)("K"))))/(1.00 color(red)(cancel(color(black)("L")))) = "6.6 atm"#