On a warm summer day, a large mass of air (atmospheric pressure 1...

Question

On a warm summer day, a large mass of air (atmospheric pressure 1.01 * 10^5 Pa) is heated by the ground to 26.0°C and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why?) Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 0.850 * 105 Pa. Assume that air is an ideal gas, with g = 1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C° per 100 m of altitude, is called the dry adiabatic lapse rate.)

Accepted Answer

Hey everyone in this problem, we have a hot air balloon at pressure equal to the atmospheric pressure. Going to use a heater to raise the temperature of the air in the balloon. Okay, so we're gonna have a temperature of 95°C. When we're at ground level. Okay, with pressure one atmosphere, the balloon is going to be allowed to rise in the atmosphere. And the process taking place in that hot air balloon, we can assume or consider to be idiomatic. Alright, so what we're asked to do is we are asked to determine the temperature of the air in the balloon. Okay, when the surrounding pressure is 0.75 atmospheres were told that we can treat the air like an ideal gas okay? With gamma equals 1.4. Alright, and we can ignore the effects of the balloon on the enclosed air. Alright, so, we're told that we can consider this is idiomatic. Okay, when we hear that, we think no heat transfer. Okay, recall that means that this process has no heat transfer. Okay, so our queue is going to be zero. All right, let's write down what we're given. Okay, so, we're given first temperature of 95°C Which is gonna be 368.15 Calvin. Were given the corresponding pressure at this temperature of one atmosphere. Okay, one atmosphere Converting this into pascal, we get 1.013-5 times 10 to the five pascal. Were given a second pressure 0.75 A. T. M. Okay, so we have P two is equal to 0.75 A. T. M. Okay, which is going to be 0.75 times 1.013-5 times 10 to the five pascal Which is equal to 7. times 10 to the four pascal. And what we're asked to find is the temperature at this corresponding pressure. So we're asked to find T. two. Okay. All right, so we're asked to find a temperature When we have idiomatic process, let's recall that we have an equation with temperature given by the following T one. V one to the gamma minus one Is equal to T two. V 2 to the γ -1. So we have T. One. We have gamma we want to find T. Two. But unfortunately, we don't have any information about the volume. V one or V two. Okay, so we can't use this equation as it is. But let's try to think about how we can write volume in terms of other quantities that we have, like the pressure. Okay, no, let's look and notice that we are told that we can treat this as an ideal gas. Okay, when we have an ideal gas, the ideal gas law applies, which says that P V. Is equal to N R. T. This means that we can write any volume as N R. T divided by the pressure P. So we can do this for the first quantities and we can do this for the second quantities. Okay we can write v. One in this form and V2 in this form. Okay because we know the pressure, we know the temperature, we're told the end. Okay let's go ahead and do this. And so I think I said we were told the end we aren't told the end but you'll see in a minute that that's okay. Alright so we have T. One. Okay times V. One. If we write V. One according to our ideal gas law we can write V. One as N. R. K. The gas constant times T. One. The corresponding temperature divided by P. One, the corresponding pressure and this is all to the exponent gamma minus one. Okay we can do the same on the right hand side. So we get T. Two V. two which is gonna be n. R. T. To overpay to. Okay, according to the gas law, ideal gas law, all to the gamma minus one. Alright so now we've written this in terms of quantities that we know and what we'll see is that we have N. R. To the exponent gamma minus one on both sides. We can divide out that term. Okay and those go away so now we don't have to worry about what that end value is. Okay so we're gonna have T one times T. One to the gamma minus one Over P to the γ -1. This is going to equal to T. two times, teach you to the gamma minus one. All divided by P two to the gamma minus one. Okay let's give herself some more room. Still know when we're multiplying with the same base. We can add exponents case. We have exponents one plus gamma minus one. So it's just gonna be exponent gamma. So we're just left with T. One. To the gamma divided by P. One to the gamma minus one is equal to T. Two to the gamma divided by P two to the gamma minus one. Okay so now for our automatic process we've written a relationship between the temperature and the pressure using the ideal gas law. Okay. And now we can go ahead and find t to that value that we are looking for. Alright so substituting in our values T. one. We had as 368.15 Calvin To the gamma. Which is 1.4. Given in the problem. Okay we had our pressure 1 1. 25 times 10 to the five pascal's. Okay. To the gamma minus one. So this is gonna be to the exponent 0.4 On the right hand side T. 2 to the 1.4 divided by P to 7. 375 times 10 to the four pascal's to the exponent 0.4 K. Gamma minus one. All right Working this out we get T to to the exponent 1.4 is equal to 3487.0585. Okay. How do we undo this? Well we need to take an exponent of 1/1 0.4. Okay. To get exponent one on the left hand side. So we're gonna get 3487.585. To the exponents one divided by 1.4. Okay? And this is going to give a T. Two value 339.1 Calvin in converting into Celsius. This is going to give us 66 degrees Celsius. So this is our T to this is that temperature that we were looking for. Okay. And if we go back to the top this is going to correspond with answer. B. Thanks everyone for watching See you in the next video.

Struggling with Physics?

Watch next

Heat Equations for Special Processes & Molar Specific Heats practice set