Question

A 1250 mL glass bulb contains 44.0g sample of propane gas, CO2, at 250 degrees Celsius. Calculate the pressure of this gas.

Answer 1

I don't think CO2 is a propane gas...:o Whatever.
Use the Ideal Gas Equation.
PV = nRT

Answer 2

We're solving for P. And we need to do some conversions.

Answer 3

Adding on to what he said.
\[PV=nRT\]
We don't know P (pressure) but we do know everything else. (v=volume in meters cubed, n is the amount of the substance in moles, R is the ideal gas constant, and T is the temperature in kelvins.)
Obviously they didn't give you the the amount of propane in moles so we have to convert that using factor label.
\[\frac{ 44.0g C3H8 (propane)}{ 1 } \times \frac{ 1 mole }{ 44.1g C3H8 } = \frac{ 44.0 }{ 44.1} mol C3H8\]
We are limited by significant figures to 1 3 figures. When you divide 44.0/44.1 you get 0.997732426, meaning that there are .998 moles of propane.
Now we put that in for n and get:
\[P \times 1.25L=.997mol \times0.08206 \times523K\]
(I got 523 by adding the degrees in celcius to 273)
Now, there's a problem. Pressure is not measured in liters. So we have to convert that to dm. 1 liter=1 dm, and we have 1.25 liters. So the conversion is:
\[\frac{ 1.25L }{ 1 } \times \frac{ 1dm }{ 1L } \times \frac{ .001m }{1 dm} = \frac{1.25 \times .001}{1 \times 1} = \frac{0.00125}{1} = 0.00125m^3\]
Putting all of this in, we get:
\[.00125P=.997mol \times .08206 \times 523K\]
\[.00125P=42.7886279\]
Divide out the .00125.
\[P=\frac{ 42.7886279 }{ .00125 }\]
\[P=34230.9023 Pa\]
But like I said earlier, you are limited to 3 significant figures, so the answer would be P=34200Pa.

Answer 4

At least, I think that's right. Best I could do with the information given, and the fact that I had to decide whether to believe this website ( http://www.chemguide.co.uk/physical/kt/idealgases.html) or google on how many dms there are in a meter.

Answer 5

Thank you so much! My chem professor proved it all right. Credit goes to you.