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Topic: Gas section help required  (Read 5991 times)

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Offline nortorius

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Gas section help required
« on: September 26, 2006, 07:15:22 PM »
Got a few questions in preparation of an upcoming test. I'm having trouble answering these...

1. A sample of gas is held in a 3.00 L container at 22 degrees celsius connected to a mercury manometer. The pressure of the atmosphere is 9.87 x 10^4 pascals and the mercury level in the manometer is 13.2cm higher on the open side of the manometer. What is the pressure of the gas, in torr (mmHg)?

Answer: 873 torr

2. A mixture of exactly two moles of helium (He(g)) and one mole of a hydrocarbon of formula CnH2n (g) is placed in a container with a small hole through which the gas can escape. The escaping gas is collected and was found to have 6.24 times more helium than the hydrocarbon. What is the molar mass of the hydrocarbon? What is the formula of the hydrocarbon (what is the value of "n")?

Answer: C3H6

Thanks in advance for any help provided. I pretty much just need something to go off of, and so it isn't necessary to fully answer the questions.

Incase you need info...

R = 8.314472 J k^-1 mol^-1 = 0.0820574 atm k^-1 mol^-1
g = 9.80665 m s^-2
1 atm = 1.013 x 10^5 pascals = 760 torr (mmHg)
0 degrees celsius = 273.15 K
density of mercury = 13.6 g/cm cubed

... and a pic for the first question:


Offline Yggdrasil

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Re: Gas section help required
« Reply #1 on: September 26, 2006, 07:59:26 PM »
Here are some hints to get you started:

1.  The presure will be the same at equal heights on the manometer.  So, at the height of the mercury column on the left of your figure, you have the pressure of your gas pushing on the mercury.  On the right side of the column, at the same height, you have atmospheric pressure plus the pressure of 13.2cm of Hg pushing on the column of mercury.

2.  Use Graham's law of efflusion.

Offline nortorius

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Re: Gas section help required
« Reply #2 on: September 26, 2006, 10:27:10 PM »
1. I was thinking I would use the equation Pgas = Pbar + delta P, but that wouldn't work because deltaP is in cm instead of a pressure unit. From your hint I'm guessing it's...

delta P = g x delta h x d

where g is the gravitational constant, delta h is the change in height, and d is the density of mercury. This means that the equation would look like:

delta P = (9.81 m s^-2)(13.2cm)(13.6g cm^-3)

But I still don't see where you get a pressure unit (atm/torr/mmHg) as an answer (maybe I'm just bad with conversions). Am I on the right track?

2. The wikipedia page looks like it iis on the verge of answering my question, but in the equation:

M1 = M1Rate21/Rate22

What do the numbers 2,1 and 2,2 beside the rate mean? Thanks again.

Offline enahs

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Re: Gas section help required
« Reply #3 on: September 26, 2006, 11:34:45 PM »


2. The wikipedia page looks like it iis on the verge of answering my question, but in the equation:

M1 = M1Rate21/Rate22

What do the numbers 2,1 and 2,2 beside the rate mean? Thanks again.

#2)
The superscript 2 (up top) is squared. The subscript (lower) means gas #2. Same for 1. On the wiki page H2 is 1 and O2 is 2.

#1)
You know if the pressure was the same the height would be the same. Find out what pressure change would make one side rise 13.2 cm of Mercury (and hint, 13.2cm converter to mm = mm of mercury = Torr).

Offline Yggdrasil

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Re: Gas section help required
« Reply #4 on: September 27, 2006, 12:57:46 AM »
delta P = (9.81 m s^-2)(13.2cm)(13.6g cm^-3)

In general this is a good formula to use to find the pressure exerted by a column of fluid.  Try converting all the units so that all distances are in meters and all masses are in kilograms.  Then recall that:

1 Pascal (Pa) = 1 N m-2 = 1 kg s-2 m-1
and
101.3 kPa = 760 torr.

However, as enahs mentioned, in the case of mercury, 1 cm Hg = 10 mm Hg = 10 torr.

Offline nortorius

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Re: Gas section help required
« Reply #5 on: September 27, 2006, 07:48:08 AM »
Got them both, thanks a lot guys.

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