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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Alain12345 on January 22, 2006, 07:48:31 PM

Title: Gas Law Questions
Post by: Alain12345 on January 22, 2006, 07:48:31 PM
I solved a couple of gas problems that my teacher gave me, but I don't know if I did them right. The questions are:

A 0.19 g sample of N2 gas is collected over water at 27*C and 96.9 kPa. What would be the volume of the dry gas at STP?

I got 0.151 L

The volume of 2.37 g of oxygen collected over water at 25*C was 1.92 L. What was the barometric pressure at the time of the experiment?

I got 92.43 kPa.

Thanks
Title: Re:Gas Law Questions
Post by: mike on January 22, 2006, 08:09:07 PM
Quote
A 0.19 g sample of N2 gas is collected over water at 27*C and 96.9 kPa. What would be the volume of the dry gas at STP?

I got 0.151 L

at the given temperature and pressure I get 0.17L and at STP I get 0.15L so I suppose you are right..
Title: Re:Gas Law Questions
Post by: Borek on January 22, 2006, 08:09:49 PM
Quote from: Alain12345 on January 22, 2006, 07:48:31 PM
I got 0.151 L

0.15L here, look at significant digits.

Quote
I got 92.43 kPa.

Hmmm, I got over 3kPa more.

Edit: Oh, well, mike was 42 seconds faster.
Title: Re:Gas Law Questions
Post by: Byrne on January 22, 2006, 08:14:08 PM
First problem:

First you need to determine the quantity of nitrogen gas:

n = m / M
= 0.19 g / 28.0 g/mol
= 6.786 x 10-3 ---> carry extra significant figure for intermediate calculation

Now, using the Ideal Gas Law:

PV = nRT
V = nRT / P
=  [(6.786 x 10-3)(8.31 kPa.L.mol-1.K-1)(300 K) / 96.9 kPa
= 0.175 L

Now use the combined gas law to determine the volume at STP:

P1V1 / T1 = P2V2 / T2
V2 = (96.9 kPa)(0.175 L)(273 K) / (300 K)(101.3 kPa)
= 0.15 L

OR you could use the following mole equation when working at STP:

VSTP = n x VM
= 22.4 L/mol x 6.786 x 10-3
= 0.15 L

You should be expressing this answer as 0.15 L, correct to two significant digits.  You cannot have an answer with more accuracy than one of the parameters involved in the calculation.

The second problem is just a matter of using the Ideal Gas Law once you determine the quantity of oxygen.

n = m / M
= 2.37 g / 32.0 g/mol
= 0.07406 mol ---> carry extra significant figure for intermediate calculations.

P = nRT / V
= 95.5 kPa

I'm getting a slightly different answer for the second problem.
Title: Re:Gas Law Questions
Post by: Alain12345 on January 22, 2006, 08:28:19 PM
I'm not really concerned about the significant digits. As long as i got the right answer, that's all that matters. I think your answer might be different because you didn't take into account that it was collected over water. The pressure of the water vapour is 3.17 kPa, so you have to subtract that from 95.5 kPa.
Title: Re:Gas Law Questions
Post by: mike on January 22, 2006, 08:40:11 PM
Quote
I'm not really concerned about the significant digits. As long as i got the right answer, that's all that matters.

not true, the answer is only right if it is quoted to the correct significant figure.
Title: Re:Gas Law Questions
Post by: Alain12345 on January 22, 2006, 09:39:37 PM
I know that, but this is just for practice. I wasn't paying attention to the significant digits when I was wrote down the answer. I just care about the process and if I'm doing it right.
Title: Re:Gas Law Questions
Post by: AWK on January 23, 2006, 01:17:51 AM
When collecting gas over water, you should take into account a partial pressure of water also. Otherwise an error of calculation is about 3-4 % in this case.
Title: Re:Gas Law Questions
Post by: Byrne on January 23, 2006, 11:58:47 PM
Quote from: Alain12345 on January 22, 2006, 09:39:37 PM
I know that, but this is just for practice. I wasn't paying attention to the significant digits when I was wrote down the answer. I just care about the process and if I'm doing it right.

If you care about the process, it would help if you actually showed it rather than just providing an answer  ;)