Calculate the mass of propane that has been used

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SUMMARY

The discussion focuses on calculating the mass of propane used from a cylinder with a height of 1.00 m and an inside diameter of 0.120 m, initially filled to a gauge pressure of 1.30 x 107 Pa at 22.0°C. Using the ideal gas law (pV = nRT), the initial number of moles calculated was 60, leading to a total mass of 2643.64 g. After the pressure dropped to 2.50 x 105 Pa, the mass of propane remaining was calculated to be 50.84 g, resulting in a mass difference of 2592.8 g. The calculations were confirmed to be correct, with a note on the importance of significant figures.

PREREQUISITES
  • Understanding of the Ideal Gas Law (pV = nRT)
  • Familiarity with gauge pressure and absolute pressure concepts
  • Basic knowledge of significant figures in calculations
  • Ability to perform unit conversions and volume calculations for cylinders
NEXT STEPS
  • Review the Ideal Gas Law applications in real-world scenarios
  • Learn about gauge pressure versus absolute pressure
  • Study significant figures and their importance in scientific calculations
  • Explore volume calculations for various geometric shapes, particularly cylinders
USEFUL FOR

This discussion is beneficial for students in chemistry or physics, particularly those tackling gas laws and thermodynamics, as well as educators looking for practical examples of gas calculations.

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Homework Statement



A cylinder 1.00 m tall with an inside diameter of 0.120 m is used to hold propane gas (molar mass: 44.1 g.mol-1) for use in a barbeque. It is initially filled with a gas until the gauge pressure is 1.30 x 10^7 Pa and the temperature is 22.0⁰C. The temperature of the gas remains constant as it is partially emptied out of the tank, until the gauge pressure is 2.50 x 10^5 Pa. Calculate the mass of propane that has been used.

Homework Equations



1.. pV = nRT
2.. p1V1 / T1 = p2V2 / T2

The Attempt at a Solution



I am really wrong here but just need a guiding hand...

The volume of the cylinder is 0.11309733m^3
From Eq 1 I get the number of moles = 60 then total mass = 60 x 44.1 = 2643.64g
Then did this again for pressure 2 (2.5 x 10^5 Pa) to get = 50.84g
Difference being 2592.8g of gas used but this seems large.

I then looked at Eq 2 and I got V2 = 0.588m^3
Not really sure where to go with that...
 
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Second equation assumes number of moles of gas have not changed - so it is of no use, some of the gas was burnt.

Your first approach looks OK, just check your math again. Mass difference is much larger than you think. Seems like you have problems with position of the decimal point.
 
Hi Borek,

I typed that wrong! It should be 0.011309733m^3

The final value of 2592.8g was calculate from that figure. So is this what you get?
 
Obviously when it comes to calculating volume of cylinder I am not better than you

Your result is OK, just watch significant figures.

--
 
LOL, thanks heaps Borek.
 

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