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39 Boyle's Law explores the effects of pressure on the volume of an ideal gas_pressure of a gas in a closed container is the container (assuming constant tempe...

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39 Boyle's Law explores the effects of pressure on the volume of an ideal gas_pressure of a gas in a closed container is the container (assuming constant temperature).to the volume of the gas in0 of 0.25 points earned 2 attempts remainingBoyle converted his observation into mathematical formula_ Which equation best represents the relationship between the pressure (P_ and the volume (V) of the gas before and after compression?0 of 0.25 points earned 2 attempts remainingAssume the initial vol

39 Boyle's Law explores the effects of pressure on the volume of an ideal gas_ pressure of a gas in a closed container is the container (assuming constant temperature). to the volume of the gas in 0 of 0.25 points earned 2 attempts remaining Boyle converted his observation into mathematical formula_ Which equation best represents the relationship between the pressure (P_ and the volume (V) of the gas before and after compression? 0 of 0.25 points earned 2 attempts remaining Assume the initial volume is 5.2 L at 0.0500 atm and the final volume is 2.00 Lj Calculate the final pressure in the container in atm: 0 of 0.25 points earned 2 attempts remaining



Answers

(III) A 0.5 -mol sample of $\mathrm{O}_{2}$ gas is in a large cylinder with a movable piston on one end so it can be compressed. The initial volume is large enough that there is not a significant difference between the pressure given by the ideal gas law and that given by the van der Waals equation. As the gas is slowly compressed at constant temperature (use 300 $\mathrm{K} )$ , at what volume does the van der Waals equation give a pressure that is 5$\%$ different than the ideal gas law pressure? Let $a=0.14 \mathrm{N} \cdot \mathrm{m}^{4} / \mathrm{mol}^{2}$ and $b=3.2 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol}.$

All right In this problem, we were asked to look at the ideal gas law in the vendor roll equation of state for a scuba tank. So this is a four part problem. I've showed the 1st 3 parts, uh, in part they were just that were asked if 2300 leaders of air are brought into a tank. That's only 12 leaders large at a pressure of initially at a pressure of one atmosphere, and the temperature is 20 degree Celsius throughout the process. Show that the amount of moles in the tank is gonna be 96. Mole weaken. Do have the ideal gas law. Pretty simply, Part B then asked to use the ideal gas law where we know the number of moles in the tank and its temperature toe predict pressure in Part C were asked to use the venerable equation to predict a pressure in the tank where we assumed about some values. The parameters for air of aid is 0.1373 India's 3.7 to times 10 to the negative five, and finally, in park. Do you were asked to compute to confirm that the percent error is about 3% between these two. So I start us off for Friday. Brass used ideal gas law, which is PV equals, and Artie, Where are is the gas constant and number moles Piece pressure V is volume and his temperature. We know all these things have, except for end, which were asked to confirm so. And it would just be P V over R T Pressure in Pascal's is about one times 10 to the five we're giving a volume of which ends up being 2.3 meters. Tubed. Gascon sins always 8.314 and units are jewels. Permal, Calvin off and our temperature just 20 degrees is to 93. Calvin, remember, we have to convert to Calvin and Pascal's. Keep our units consistent. We're left with 95.64 mole, which is approximately 96 malls, which is what we were asked to confirm. That's it. That's not so bad for Barbie. We once again used ideal gas, Law said. Now we need to find pressure. So PV goes and Artie, we're given the number of moles in the tank in this case, and the temperature and the volume of the tank So pressure is just going to be put it over there. So pressure is just a party over V. So we go, we have to convert these volumes two meters cubed the point tank. And you just gave this 0.12 number. Moles, we assume, is 96. That's given in part B. Once again, the gas constant 8.314 and a temperature of 2 93 Calvin or 20 threes. All right. And we end up with about 1.9 times 10 to the seven Pascal Cool. Which is about 190 atmospheres. Remember, one atmosphere is about 10 to the five. Pascal. Cool. So you can see that this air in the tank is quite pressurize is almost 200 times the pressure. Uh, regular. All right, So I'm gonna go to another page for valuable equation, cause usually pretty large. Let's do that. All right, So remember the van rolls equation p. The predicted pressure for the mandibles equation is Artie over via Veran. My say it must be I always mess that up, and then it has a second term minus a over the over and squared Great and so we can expand this a little bit to make it a little bit easier to calculate things. This is equivalent to an arty over V minus N B minus and squared over V squared. Okay, so we're gonna use the exact same values as in part B for the volume. So just write that if volume is ah 0.12 meters cubes temperature is to 93 as before, Kevin and we have our values A and B from before. What was it? It was, uh, writer's point 1373 And I'm just gonna take the units, officer kind of messy and B is 3.72 times 10 to the negative five. We could put all these values in and compute p, which comes out to be about 1.9 times tender. That seven Pascal. All right, So far, so good. Cool. So, finally, our last part is, all we have to do is compare these two values. Um, and because they're so close to each other, actually, to use more precise values, get this percent air. I've just included the approximation toe like one or two significant figures, but we have to do a little bit better than that. I'll include the more precise values here. So we have a pressure from the ideal gas law as the pressure from the Vander Wal's equation or the pressure on the valuables equation. We're just getting a percent air. Here's were just comparing the pressure from Parsi and the pressure from party. All right, so now I'm gonna quote you some more precise values of these. We can actually get a good percent air. You have 1.9488 captain of the seven Pascal. That's ideal gas law. And 1.89 58 times 10 to 7. So I just listed these as 1.9 each in part B and C, and that's sufficient. But now we have to be a little bit more precise. Okay? And once we calculate this, we end up with 2.796% which is approximately 3% which is what we were asked to D'oh! And that's it. We've confirmed it. So the Van der Waals equation ideal gas law both give you pretty close estimates for pressure inside scuba tank. They're slightly off, but ah, pretty consistent with each other. So that's it. Just practicing using those two equations estate for the gas

In the institution. It is told data sample of an ideal guest was initial pressure and volume is given to us and it is told that the guesses I saw thermally taken to a pressure to note by two. And from there it is an a diabetic process where pressure is be not by four. Now we have to find the final volume. So first of all we have given debt initial pressure that is proven his peanut and initial volume is we're not even as equals two. We note for I saw thermal process. I can say that that is P even everyone is equal to P to veto here everyone is peanut and we even if we note is equals two P. Two and who is given to us that is we note by two And this is we too were given that Peter is Peanut by two. From here we get the volume and I saw thermal process that is two times of we not now after that it is taken to an a diabetic process so forth. A diabetic process. I can say that P two way to to the balderrama is equal to P three V three to the power gamma. We have given the value of P three that is not by four and we have to find the value of the three. Now put all the data in the situation so pay two equals 2. Or I can simplify this equation that we three by way to to the power gamma is equals two. You too bye. P three. Now for the value of P two. P 3. So this is We three by me too. To the power gamma is equals two. This is We're not by two and this is not by four. From here we get the value of the three. There it is two to the power one upon gamma into V two. Now we have the value of the two. So this is two to the power one upon gamma and we do is two times of we note. So from here we get to and to gamma to the power gamma plus one upon gamma into we note. So this is the final volume for the first part of our given solution. Now in the second part it is told that the gas is brought back to the its initial state and it is a diabetic lee taken to a pressure of the note by two. So first of all the processes, A diabetic so far, a diabetic process. I can say that even we went to the program a is equal to p. two. We two to the power gamma. No, we can see that From here we do by we want to the power gamma is equals to be one by P two. So we get the value of the two dirties equals two. We're not to the power we're not sorry to to the power one upon gamma. This is the volume after a diabetic process. Again the process is Isil thermal. So in I saw thermal process we can see that B two, B two equals two. B three V 3. So from here we get the value of the three dirties. P two we two upon P three. Put all the data in this situation. So this comes out to be two into two. To the power one upon gamma into we note after simplifying this, we get this is two to the power gamma plus one upon gamma into we note this is the final volume so we can see that the final volume in the boat, the cases due to the power gamma plus one upon gamma into we note four both. Yes, yes. And this is our answer for this given solution. Thank you

So here we are assuming ideal gas behaviors, we can calculate the number of moles help moles is equal to P V over R T, where we have 300 atmosphere multiplied by 10 Destiny two cubed, divided by 9.98206 decimal to cube multiplied by atmosphere multiplied by Kelvin to the minus one multiplied by most of the minus one, multiplied by 293 Kelvin. Our moles here is 1 to 4 moles. In the second part, we can use the equation. P one Taiwan is equal to p two V two, where we want p two so we can rearrange. P two is equal to 300 atmosphere invited by 293 Calvin multiplied by 278 Calvin P two is equal to 285 atmosphere. So in the third party it. So we need to recall that the two assumptions for the ideal gases are considered to the fast assumption is that there are no inter molecular forces attractions between the molecules and the second assumption is that they are negligible size. So when the gases are compressed, the gas molecules will come closer to one another and There is a higher inter molecular force of attraction between the molecules due to the smaller force exerted by the molecules at the walls of the container and the pressure will decrease slightly so the size of the gas molecules will increase You too. The volume in which the gas molecules will move about. This is lower now than the volume of the container. So therefore the compressed gas do not show ideal behavior. So the value of a M B. Ah, we know those. And then we can use the conversion to find the volume. So the volume is 10 to the minus 2 m. We need to convert from Destin meters, 2 m that we can convert Pascal for the pressure to 30.4 times 10 to the six Pascal and then we have the R gas constant. That is 8.314 So we can solve for the temperature. T is equal to p. R. A n squad divided by B squared multiplied by V. Take away and be divide that by an off We got a temperature of 291.5 Calvin, So the temperature almost matches the given temperature so the number of the malls or amount of gas in the scuba cylinder is actually 115 moles and not 124 which indicates that the amount of gas or air in the cylinder is always less than the calculated value from the ideal gas equation.

Action is asking us for a final pressure based on the relationship between pressure and volume. So we know that we will be using Boyle's law, which states that pressure and volume are inversely correlated. So the equation for that is P one times V one equals P two times fee, too. And since we were solving for our final pressure, we're solving for P two, so we'll be rearranging this equation. Two p two equals p one times V one over the two. And for this, because all of the values air given to us, we can just plug those in, knowing that our initial pressure is one atmosphere. Our initial volume is 3.25 leaders, and our final volume is 2.24 leaders and solving for P two. We will find that the final pressure is 1.45 atmospheres with significant figures. And just make sure that you're using the same units for volume so that they'll cross out in the


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