How does water vapor affect the expansion of butane in a lighter?

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Water vapor significantly impacts the expansion of butane in a lighter, as the pressure differential between the lighter's interior and the atmosphere allows liquid butane to vaporize, increasing volume. The combined gas law can be applied to understand this behavior, as it relates pressure, volume, and temperature changes. When conducting experiments, it's crucial to account for water vapor's presence, as it can alter the calculated molar mass of butane. Accurate measurements of gas volume should be corrected to standard temperature and pressure (STP) to ensure reliability. Understanding these principles is essential for analyzing the behavior of gases in different conditions.
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i was interested if anyone could explain how a butane lighter can expand from liquid state into the gas state with what appears to be, more volume?

it'd be great if you could mention any laws or theories

thanks :)
 
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PV=NkT

Liquid is denser than gas.
 
thanks

any way/formula of measuring the mass when it changes?

would the combined gas law work in this instance?
 
You can measure the mass change with a scale, but the density of the liquid you can probably just google.
 
kateman said:
i was interested if anyone could explain how a butane lighter can expand from liquid state into the gas state with what appears to be, more volume?

it'd be great if you could mention any laws or theories

thanks :)
It's a matter of reduced pressure. Inside the lighter casing/reservoir, the pressure is slightly higher than atmospheric pressure. The liquid/vapor inside is satured, such that the pressure is fixed according to temperature. When a little vapor is released (because of the pressure differential between inside and out), more liquid becomes vapor to maintain an equilibrium.

Outside the lighter, the pressure is 1 atm.

See -
http://www.engineeringtoolbox.com/propane-butane-mix-d_1043.html (annoying popups on this site).
Propane is more suited to colder environments since it evaporates at -44°F (-42°C) at atmospheric pressure. Butane evaporates at 33°F (-0.5°C) at atmospheric pressure.

The vapor pressure of a mixture of the two products can be found in the table below:
 
thank-you Astronuc, that is exactly what i was after
one more quick question, i did an experiment of using a butane lighter underwater and collecting the gas in an inverted cylinder so that the butane gas displaces the water and from there finding the molar mass of butane.

Well the lighter lost 0.04 grams of butane yet there were 101ml of gas in the cylinder. I worked it out (assuming butane is an ideal gas, which its not) so that butane had a molar mass of 9.46g (which is way off).
The working is right, so what could my experimental error could have been?
 
How was the volume of the gas measured? Also, make sure volume measurements are corrected to STP.

Reweigh the lighter.

Also, was the volume dry or was water vapor present. A significant mole fraction of water vapor would drop the molecular mass.

The molar mass is ~58 g/mole.

The density of the gas is 2.52 g/l or 0.00252 g/ml, so with 101 ml, one should have 0.255 g, so one is off somewhere by a factor of about 6.
 
the volume of gas was measured by finding the amount of water left in the inverted cylinder and taking it off the total capacity of the cylinder.

water vapor was present. it was taken of the room pressure to give the butane pressure.


"A significant mole fraction of water vapor would drop the molecular mass."
how would find the volume of water vapor in the cylinder and express that mole fraction algebraically?
 

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