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Pressure/Mass Relationship

  • Thread starter octahedron
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  • #1
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[SOLVED] Pressure/Mass Relationship

1. Homework Statement

[tex]H_{2}[/tex] gas was obtained by the displacement of [tex]H_{2}O[/tex] at total pressure of 833 torr. The volume of the obtained gas was .5 Liters and the vapor pressure of [tex]H_{2}O[/tex] was 22 torr. Find the mass of [tex]H_{2}[/tex] gas.

2. Homework Equations

[tex]P_{tot} = P_{1} + P_{2}[/tex]
[tex] P_{tot} = (n_{1} + n_{2})(\frac{RT}{V})[/tex]
[tex] P_{1} = (n_{1})(\frac{RT}{V})[/tex]
[tex] P_{2} = (n_{2})(\frac{RT}{V})[/tex]

3. The Attempt at a Solution

Now here is my attempt, which I'm definitely not sure of:

[tex]P_{tot} = P_{H_{2}O} + P_{H_{2}} \Rightarrow P_{H_{2}} = 833 - 22 = 811[/tex]

Then,

[tex]n_{H_{2}} = \frac{PV}{RT} = \frac{(811)(0.5)}{(0.082)(295)} = 16.7[/tex]

And then trivially obtain the mass of hydrogen in the experiment:

[tex]m_{H_{2}} = (2)(16.7) = 33.4 g[/tex]

Is this correct? I'm not sure of how I obtained [tex]P_{H_{2}}[/tex].
 
Last edited:

Answers and Replies

  • #2
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I'm going to assume this was at a temperature of 295°K, because that's what you used in your calculations, but you didn't state this in the problem.

You calculated the pressure of the hydrogen gas correctly. The total pressure of the gas is 833 torr, with H2O gas accounting for 22 torr of that pressure. This means that the hydrogen gas is responsible for the remaining 811 torr of pressure.

The equation you used was fine, but you didn't use the correct "R" value. If you use torr for pressure, you can't use 0.0821 for the R value. Divide the pressure by 760 to convert to atmospheres, which then lets you use 0.0821 for the R value. Or you can use the proper R value for when you have a pressure measured torr, but I prefer to only have to memorize R value.

So just divide the answers you got by 760 and they will be correct.
 
  • #3
37
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Thanks nautikal!
 

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