A cylinder 1.00 m tall with inside diameter 0.120 m is used to hold propane gas (molar mass 44.1 g/mol) for use in a barbecue. It is initially filled with gas until the gauge pressure is 1.30 $\times$ 10$^6$ Pa at 22.0$^\circ$C. The temperature of the gas remains constant as it is partially emptied out of the tank, until the gauge pressure is 3.40 $\times$ 10$^5$ Pa. Calculate the mass of propane that has been used.
Thermal Properties of Matter
in this question. We have a cylinder 1 m tall with a diameter of 0.12 m that is used to hold propane gas. Ah, for a barbecue. It is initially filled with gas until the gauge pressure is 1.3 times 10 of the six Pascal's at 22 degrees Celsius. And we're told that the temperature remains constant as it is partially emptied out of the tank until the gauge pressure is 3.4 times 10 to the five Paschal's. So the question is asking us to calculate the mass of the proton propane that has been used, presumably by the barbecue. So let's go ahead and start out by writing out some givens because there's quite a lot of information that's given in this question. So we have the heights of the propane tank is 1 m. Then we were told that the diameter is 0.12 m. So what I'm going to do is just divide that in half to get the radius, and then we're told that the molar mass of the proton propane is 44.1 g per mole. Oh, and then we're given some gauge pressures. So the initial gauge pressure is 1.3 times 10 to the six Paschal's, and the final gauge pressure is 3.4 times 10 to the five Paschal's. And let's see. Ah, the temperature is 22 degrees Celsius. For most of these calculations, we need the temperature to be in Calvin's. So let's just go ahead and add on the 2 73 right now. So that's going to give us to 95. Calvin's so looking at this information, this is kind of a hint that we can calculate the volume, the height in the radius of a cylinder. We can calculate the volume. Um, we've got some pressures and we know the temperature. So this is kind of hinting towards using the ideal gas law PV equals and R T. And what we can do is use this equation to calculate the number of moles before the propane starts to be used, and then after as well. And then we just subtract them and we'll get the number of moles that, um, that have been used overall, and we can convert that to a mass. So let's follow that plan. Um, and the one thing you want to be careful with here. Uh, of course, we're going to rearrange this so and is equal to P V over R T theme P does have to be the absolute P. So the absolute pressure, um, experienced by the gas. So it's not just the gauge pressure that were given in the question, but we also have to add in the atmospheric pressure so we can fix that by adding 1.13 times 10 to the five Paschal's to each of these pressures that we're going to use. So for the initial, um, Mueller or the initial amount of moles, um, we're gonna use that initial pressure. Let's see. So adding on the atmospheric pressure to that would give us 1.4 zero times 10 to the five Sorry, 10 of the six Paschal's, then for the volume. Um, I'm going to use the formula pi r squared times h. So once we do that, we get a volume of 0.113 meters cube. So that's the volume I'm going to use. And then the are the standard are 8.31 approximately, and then we use our tea. So once we do that, we find a an amount of moles. 6.46 moles. Okay, we're going to do that again, basically with all of the same variables. Um, but now we're going to use that pressure. Um, that second pressure. So again, you want to make sure that you add on the atmospheric pressure. That's especially important this time. So that's 4.41 times 10 to the five pass scales. Ah, we can use the same, um, the same volume for this cylinder because the propane cylinder, that's not going to change size, harsh the same, obviously. And we were told that the temperature stays a constant, so we're going to go ahead and use to 95 again. And so once you do that calculation, you should get a molds of 2.3 moles. So left, um, in the tank is 2.3 moles. We started with 6.46 moles. So the amount that escaped the amount that was used by the barbecue is en one minus end to so that will give us an amount that has been used of 4.43 moles. And remember in the question we were given the the Mueller Mass. So we can use, um, equal to and times big M to find the mass of the gas that has been used. And I'm just gonna leave that I am in grams. So our final answer will be in grams. So once we do that, we see that about 195 g has been used or the equivalent of 0.195 kg. So this is our final answer for this particular question here.