Physics Help Forum Ideal Gas Equation - Problem regarding mass

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 Apr 7th 2017, 10:29 AM #1 Junior Member   Join Date: Apr 2017 Location: London Posts: 5 Ideal Gas Equation - Problem regarding mass Hi people, I needed some assistance with this problem I came across that involved the use of the ideal gas equation. Q. A gas cylinder has a volume of 20 litres (20*10^3 m^3). It contains air at a temperature of 17 degrees Celsius and an excess pressure of 3.0*10^5 Pa above the atmospheric pressure 1.0*10^5 Pa. Calculate the mass of air in the cylinder, given that the density of air at STP is 1.3 kg/m^3. A. 98*10^-3 kg. My attempted solution at this was: PV = nRT By using the equation above we can work out the number of moles "n". So rearranging to make n the subject of the formula. n = PV/RT Therefore n = (4*10^5 N/m^2)(20*10^-3 m^3)/(8.31 J/kg*K)(290 K) We get n to be 3.32 (correct to 3 s.f.) Since n is also = m/Mr Where m is the mass in kg and Mr is the molar mass. m = n*Mr = (3.32)(0.029) = 0.09628 kg As you can you can see I failed to arrive at the correct answer, and I am unsure why we have been provided the density of air at STP? I would appreciate any help, thank you in advance!
 Apr 7th 2017, 05:21 PM #2 Senior Member   Join Date: Jun 2010 Location: NC Posts: 388 I think you're right. Don't know why the density was included. Avoid algebra with this equation. Using the Gas Equation| THERMO Spoken Here! Victor Regnault | THERMO Spoken Here! Good Luck with this... JP Attached Thumbnails
 Apr 8th 2017, 04:18 AM #3 Junior Member   Join Date: Apr 2017 Location: London Posts: 5 Oh alright, thank you for responding I appreciate it
 Apr 8th 2017, 03:42 PM #4 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 926 I don't think this is right at all. There are several ways to do this but none involved switching to the molar equation. Where did the question tell you the molar mass of air? The density was given because it tells you how much is in a given volume, but at a different temperature and pressure from the one the gas is at. So you must either correct this density to the new temperature and pressure or reduce the given values to STP. Then you can use the information to obtain the mass of gas. P1V1/T1 = P2V2/T2 Since you are given P1, P2, T1 and T2 you can correct 20 litres of gas at 4 atm and 17C to V2 litres of gas at STP and multiply by the given density.in kg/litre skengdo410 likes this.
 Apr 9th 2017, 06:05 AM #5 Junior Member   Join Date: Apr 2017 Location: London Posts: 5 @studiot I don't understand how I can make any use of V2? Also V2 is the volume at which the air at STP is occupying right?
 Apr 9th 2017, 06:19 AM #6 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 926 Start You have V1 = 20 litres of gas in a sealed cylinder at P1, T1. (given) This is a fixed mass of gas. (Theoretically) Allow this mass of gas to expand to V2 at STP so it is now at 273K and 1atm. Calculate V2 for this state. You are given the density of the gas in this state. Using the density and your calculated V2 you can calculate the mass. There are many online calculators using this method as it is a common requirement to reduce or correct to STP. skengdo410 likes this.
 Apr 9th 2017, 06:30 AM #7 Junior Member   Join Date: Apr 2017 Location: London Posts: 5 @studiot Bro, thank you so much for the explanation, I have arrived at the correct answer So this means that PV = nRT can only be applied for fixed masses right? And PV = mrT is to be used when the masses change?
 Apr 9th 2017, 07:03 AM #8 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 926 You can apply PV = mRmT if you use the correct 'universal' gas constant, Rm in this case. 'm' is the mass of gas not the number of moles. In this case it is not the constant that uses moles, but the constant that uses mass in its definition. Good textbooks will offer R in many different units as we often need to swop.

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