Calculating the volume of gas in the same canister from 2 locations.

[LAB]

I have a canister with a capacity of 8334cm3, if the canister is sealed at ambient atmospheric pressure (1016.5hPa, 25.6 deg/C) at one location then transported thousands of kilometres where the ambient pressure and temperature (1014.5hPa, 20.5 deg/C) is different.

I then connected my canister to an inverted measuring cylinder over a water bath to physically measure the difference in the volume of gas in the canister, I got a reading of -700mL (started at 1000mL finished at 300mL).

How do I mathematically calculate the difference in volume of gas in the canister in mL's ?

 

[Nessdude14]

You'd use the ideal gas law, PV=nRT
P is pressure, V is volume, n is number of molecules of substance in mols, R is the ideal gas constant (8.31 J/mol K) and T is Temperature (in Kelvin).

From the parameters you measured, you can rearrange the equation to determine the number of mols of substance in your container.
n=(PV)/(RT)

If you then want to know the volume of that container for a given pressure and temperature, you can again rearrange the equation:
V=(nRT)/P
and as long as the container remains sealed, n should remain constant.

However, this does assume that no gas was able to permeate your container (which is nearly impossible in reality).

 

[LAB]

Thank you Nessdude14,

After doing the calculation according to the Ideal Gas Laws sugested, the volume of gas at location 2 with lower temperature and pressure equals ~154mL less than location 1, but when I physically measured it; the result was 700mL less.

Can anyone sugest why this is happened? or did I calculate it in correctly?

 

[CWatters]

How was the gas transported and in what?

If by air then gas might have leaked out due to the reduced cabin/hold pressure in flight. As you descended the increasing pressure might have resealed the container (eg gas inside now stuck at the lower cabin pressure). Hence much lower volume when measured.

If the canister is a special gas bottle then this is unlikely to be the cause - unless it wasn't sealed properly?

 

[pmacias]

To mathematically calculate the difference in volume of gas in the canister, you will need to use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature.

First, we need to convert the given pressures to their equivalent units in pascals (Pa), as the ideal gas law equation requires pressure to be in Pa. So, 1016.5 hPa is equivalent to 101650 Pa, and 1014.5 hPa is equivalent to 101450 Pa.

Next, we need to calculate the number of moles of gas in the canister at both locations. This can be done using the formula n = PV/RT, where P is pressure, V is volume, R is the ideal gas constant, and T is temperature in Kelvin (K). So, for the first location, n1 = (101650 Pa)(8334 cm3)/(8.314 J/mol*K)(298.6 K) = 0.349 moles. Similarly, for the second location, n2 = (101450 Pa)(8334 cm3)/(8.314 J/mol*K)(293.5 K) = 0.347 moles.

Now, we can use the ideal gas law equation to calculate the volume of gas at the second location, V2 = (n2RT2)/P2, where T2 is the temperature at the second location in Kelvin. So, V2 = (0.347 mol)(8.314 J/mol*K)(293.5 K)/(101450 Pa) = 8,334.17 cm3. This is the theoretical volume of gas at the second location, assuming the canister maintained the same pressure and temperature conditions as the first location.

To calculate the difference in volume between the two locations, we simply subtract the theoretical volume at the second location from the actual volume measured in the inverted measuring cylinder. So, the difference in volume would be 8,334.17 cm3 - 300 cm3 = 8,034.17 cm3, which is equivalent to -8034.17 mL. This means that the volume of gas in the canister decreased by approximately 8034 mL due to the change in pressure and temperature.

It is important to note that this calculation assumes ideal gas behavior, which may not be completely