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Topic: Gases, kPa and Pa  (Read 4892 times)

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Offline Olya

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Gases, kPa and Pa
« on: November 29, 2006, 12:54:40 AM »
Could somebody please explain to me where my mistake lies?


Problem:
A barometer is filled with triethylene glycyol. The liquid height is found to be 7.39 m when the atmospheric pressure is 757 mmHg. What is the density of triethylene glycol?

Solution:

1)Pressure(Pa) = d (kg/m^3) x h (m) x g (m/s^2)

d = P/ h x g

2) P = 757 mmHg

757mmHg x 101325 Pa/760 mmHg = 100 925 Pa

3) d = 100 925 Pa/7.39 m x 9.80665 m/s^2
      = 1.39 kg/m^3

Answer in the book = 1.39 g/cm^3....
« Last Edit: November 29, 2006, 06:01:34 PM by Olya »

Offline Yggdrasil

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Re: Gases, kPa and Pa
« Reply #1 on: November 29, 2006, 02:54:42 AM »
I think the books answer should be in units of g/mL.  If you check your calculations then convert to g/mL (1mL = 1 cm^3) you should get 1.39 g/mL.

Offline Olya

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Re: Gases, kPa and Pa
« Reply #2 on: November 29, 2006, 06:01:57 PM »
I think the books answer should be in units of g/mL.  If you check your calculations then convert to g/mL (1mL = 1 cm^3) you should get 1.39 g/mL.

sorry, they are in g/cm^3...but why?

Offline Dan

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Re: Gases, kPa and Pa
« Reply #3 on: November 29, 2006, 06:18:23 PM »
3) d = 100 925 Pa/7.39 m x 9.80665 m/s^2
      = 1.39 kg/m^3

You've got the wrong number out of this by a factor of 1000. It's just a calculator input error. Do it again (carefully) and then convert to g/mL.
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