A 7.25-L container holds a mixture of two gases at 37 °C. The partial pressures of gas A and gas B, respectively, are 0.352 atm and 0.810 atm.

Question

If 0.220 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

it says answer should be P total=___ atm

I keep getting 1.93 and it says its wrong

Answer 1

Hello, Aundrea,

Interesting problem. The approach I took was to use the ideal gas law to first calculate how many moles are represented by the two gases, A and B, under the conditions specified. I then added the moles of the third added gas (I'll call it C) and calculate the resulting pressure, also with the ideal gas law.

PV=nRT

I want n, total moles, for the two gases A and B, so rearrange:

n = PV/RT

R = 0.08206 L*atm*K^-1mol^-1

P = 0.352 atm + 0.810 atm = 1.162 atm

T = 310.2 K (add 273.2 to convert C to K)

V = 7.25L

Enter the data and make sure to cancel units where possible. We should wind up with only moles. I find gases A and B amount to a total of 0.331 moles.

Then add the 0.220 moles of gas C, for a total of 0.551 moles of gases.

No let's use this number of moles in the ideal gas law. Now we want to find pressure.

P = nRT/V

We use the new value for moles, but the others remain the same.

P = (0.551 moles)(0.08206 L*atm*K^-1mol^-1)( 310.2 K)/(7.25L)

Hmmmmmm . . . . . .

I also get 1.93 atm!

I don't know why that is incorrect. Perhaps it is only expecting 2 sig figs? The temperature ony has two, so the legal answer would be 1.9 atm. Although I hate giving up hard-won digits, that is often a problem with computer-graded answers. There is rarely a "Nice job, but suggest looking at sig figs before submitting" answer. The thoughtless response is generally "INCORRECT."

Please let us know what the "correct" answer is once you find out. I'm very curious.

Bob

Answer 2

There are a couple of ways to solve this problem. One way is to find the pressure contributed by the third gas alone, and then add that to the pressure of the other two gases. We can do this by using the ideal gas law,

PV = nRT for gas #3, and solving for P (pressure)

P = nRT/V = (0.220)(0.0821)(310) / 7.25 = 0.772 atm

Total pressure = 0.772 atm + 0.352 atm + 0.810 atm = 1.93 atm which agrees with your answer.

Checking this by a second method:

PV = nRT and solve for moles of A and B

n = PV/RT

For gas A: n = (0.352)(7.25) / (0.0821)(310) = 0.100 moles of A

For gas B: n = (0.810)(7.25) / (0.0821)(310) = 0.231 moles of B

moles of C = 0.220

Total moles = 0.100 + 0.231 + 0.220 = 0.551 total moles of gas present

Solving for pressure: P = nRT/V = (0.551)(0.0821)(310) / 7.25 = 1.93 atm