Alain12345 - 22-1-2006 at 13:18
I solved a gas problem that my teacher gave me, but I don't know if I did it right. The question is:
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, but I'm not sure if it's right. Can someone correct me if I'm wrong? Thanks.
EDIT:
Sorry, I've got one more...
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.
[Edited on 22-1-2006 by Alain12345]
solo - 22-1-2006 at 14:22
Her is a step towards checking your answer......as I can remember......solo
P1V1/T1=P2V2/T2
one atmosphere =101.3kPa
Given:
.19gr of N2 .19/14x2=.19/28=.00678 moles
Temp 25C=298.5K
96.9kPa
Note: similar problem
Example #2 - This next problem uses two gas laws in sequence. It involves using Dalton's Law of Partial Pressures first, then use of the Combined Gas
Law. The explanation will assume you understand Dalton's Law. These two laws occuring together in a problem is VERY COMMON.
1.85 L of a gas is collected over water at 98.0 kPa and 22.0°C. What is the volume of the dry gas at STP?
The key phrase is "over water." Another phrase to look for is "wet gas." This means the gas was collected by bubbling it into an inverted bottle
filled with water which is sitting in a water bath. The gas bubbles in and is trapped. It displaces the water which flows out into the water bath.
The problem is that the trapped gas now has water vapor mixed in with it. This is a consequence of the technique and cannot be avoided. However, there
is a calculation technique (Dalton's Law) that allows use to remove the effect of the water vapor and treat the gas as "dry." For this example, we
write Dalton's Law like this:
We need to know the vapor pressure of water at 22.0°C and to do this we must look it up in a reference source.
It is important to recognize the Ptot is the 98.0 value. Ptot is the combined pressure of the dry gas AND the water vapor. We want the water vapor's
pressure OUT.
We solve the problem for Pgas and get 95.3553 kPa. Notice that it is not rounded off. The only rounding off done is at the FINAL answer, which this is
not.
Placing all the values into the solution matrix yields this:
Solve for x in the usual manner of cross-multiplying and dividing.
mick - 23-1-2006 at 13:50
I mol of an "ideal gas" occupies 22.4 L at STP whether they are different gases.
I think PV=nrT
n is the number of moles
r is the gas constant, if I remember 8.12 or something.
If you have 10 molar% oxygen or methane or hydrogen with 90 molar% nitrogen or helium or hydrogen sulfide and they do not react or liquify and are not
under more than atmospheric pressure they should occupy close on 22.4 L.
mick
0.19 g of nitrogen will be the same volume at STP whatever temp and pres you collected it.
mick
[Edited on 23-1-2006 by mick]
runlabrun - 24-1-2006 at 16:12
PV = nRT
V = nRT / P
n = 0.19g / (14.01x2)gmol-1 = 0.00678mol of N2
R = 8.20578x10-2 (atm.L)/(mol.K)
T = 273.15 + 27 = 300.15K
P = 96.9kPa = 0.9563286atm
V = [0.00678 x 8.20578x10-2 x 300.15] / 0.9563286
V = 0.1746L
Yes?
And the other one...
PV = nRT
P = nRT / V
n = 2.37g / (16 x 2)gmol-1 = 0.074mol O2
R = 8.20578x10-2(atm.L)/(mol.K)
T = 273.15 + 25 = 298.15K
V = 1.92L
P = [0.074 x 8.20578x10-2 x 298.15] / 1.92
P = 0.9429atm
P = 95.53934kPa
Yes?
Check your rounding and significant figures and be sure to use the correct gas constant for the units to be cancelled which leaves you with the
correct units for the answer.
-rlr