(II) A 2.00-mole sample of N₂ gas at 0°C is heated to 150°C at...

Question

(II) A 2.00-mole sample of N₂ gas at 0°C is heated to 150°C at constant pressure (1.00 atm). Determine (a) the change in internal energy, (b) the work the gas does, and (c) the heat added to it.

Accepted Answer

Everyone in this problem, we're asked to calculate one, the change in internal energy deal to you. Two of the work done by the gas W and three, the heat added to the gas Q. If a constant pressure of one atmosphere, a 2.5 mole sample of 02 gas undergoes heating from an initial temperature of 10 °C to 200 °C. We're asked to consider the molar specific heat capacity at constant volume or 02 gas CV equal to 29.1 joules per mole P. We have four answer choices options A through D each containing a different value for these three things. So the change in internal energy, the work done and the heat added, all of those values are in joules. And welcome back to these answer choices at the end. Now, the first thing we wanna think about here is that we have a constant pressure process. OK. So because we have constant pressure, that means that we have an isobaric process. OK? So as we think about all three parts, we're gonna think about an isobaric process. All right. So let's get started with part one and for part one, we're looking for the change in internal energy delta and recall that the change in internal energy can be written as N multiplied by CV, multiplied by delta T. They were N is a number of moles CV is a heat capacity at constant volume. And delta T is a change in temperature. And we have to be very, very careful with this equation that we have CV and not C PAC P gives us a different equation. That's one that we can use for the heat. OK. So CV, in this case, we have these values. OK. So let's write out what we were given. So we have them written nicely but we have this pressure p, this constant pressure of one atmosphere, we have the number of moles. It's 2.5 we have our initial temperature T knot of 10 °C, our final temperature 200 °C. And then the CV value were given as well 29.1 jules thermal. OK. So we can go back to part one, our change in internal energy. We have the values, let's go ahead and substitute them. So, and that's gonna be 2.5 more multiplied by CV, 29.1 jewel Hermo Calvin multiplied by the change in temperature. Now, in this case, we wanna convert our temperature into Kelvin so that we can divide it with the Kelvin in the denominator to do that. We add 273 to each of them. So our final temperature 200 plus 273 gives us 473 Kelvin. And our initial temperature 10 °C plus 273 gives us 283 Kelvin. So in our equation, we have 473 minus 283 Kelvin, we can work this out on our calculator and we get that the change in internal energy delta U is gonna be 13,822 0.5 joules. And so that is part one done. We're gonna round to two significant digits. If we take a look at the answer choices for delta U, we have option A and C with 14 multiplied by 10 to the exponent three Juls, we have B and D as 9.5 multiplied by 10 to the exponent three Js. OK. So B and D we can eliminate those don't have the correct value there but A and C are options. Now we're gonna move on to part two and part two asked us for the work done by the gas. Now, the work done because we have an isobaric process or a constant pressure process. Recall that the work done can be written as that pressure multiplied by the change in volume. OK. So we can write this as a pressure multiplied by the final volume minus the initial volume. We know the pressure P, but we don't know these volumes, the final and the initial volume. So what we're gonna do is we're gonna go ahead and we're gonna use the ideal gas law or call the ideal gas law tells us PV is equal to N RT. Or if we're looking for the volume V, like in this case, it's gonna be equal to N RT divided by P. So we can use this equation for both the initial volume and the final volume. We have all of the other information. So our initial volume VNO is going to be equal to the number of moles. 2.5 multiplied by the gas constant, 8.314 joules per mole Calvin multiplied by the initial temperature. 283 Calvin, all divided by the pressure, which we're told is one atmosphere or 101,325 pascal. When we work this out, we get an initial volume V not of about 0.05805 m. Cute. Now we can do the exact same thing with the final volume we substitute in those values 2.5 mil multiplied by 8.314 joules per mole Calvin. The temperature has changed. So now we multiply by 473 Calvin and we divide by that same pressure because the pressure doesn't change from the initial to the final point 101,325 pascals. And we get this final volume. VF how about 0.09 7 0 2 7 m cubes? So the volume has increased. OK. So we have our pressures now or our volumes. Sorry, we've calculated our volumes. We can get back to our equation for work. We're gonna have the pressure 101,325 pascals multiplied by these volumes we just found. So we have 0.097027 m cubed minus 0.0 5805 meters. Cute. We work this out and we get a work about 3949.34 45 jewels. OK? So that's part two that we needed to find those volumes first using the ideal gas law. But then we were able to calculate the work using P delta V. If we take a look at our answer choices again, rounding to two significant digits. We had already narrowed it down to A and C. Both of those have that the work is 3.9 multiplied by 10 to the exponent three joules. OK. So both of those could still be correct. And we're down to this final section. Now, for part three, we mentioned this before, but we could write the heat Q as N multiplied by CP multiplied by delta T. The problem here is that we don't have this CP value and remember we were given CV. So always double check, is it CV or is it CP and make sure you're using it in the correct equation. And so we can't go about it this way. But what we do know is that the heat Q is related to the work and the change in internal energy, we already calculated those two. So we can write that delta U is equal to Q minus W or alternatively, the heat Q is equal to the change in internal energy delta U plus the work W adding together the two values we just found hun 0 13,822 0.5 joules plus 3949.344525 jewels. When we work this out on our calculator, what we find is that the heat absorbed Q is about 17,771 0.8445 jewels. And that is the final answer for part three. Again, we're gonna round to two significant digits we're gonna compare with what we found in the answer choices or what we see in the answer choices. Option A has for Q 13 multiplied by 10 to the exponent three jewels. That's not what we found. So we can eliminate that. But option C has the correct value of 18 multiplied by 10 to the exponent three jewels. So answer C is going to be the correct answer here. Change in internal energy was 14 multiplied by 10 to the exponent three jewels. The work done 3.9 multiplied by 10 to the exponent three jewels and the heat absorbed 18 multiplied by 10 to the exponent three jewels. Thanks everyone for watching. I hope this video helped see you in the next one.

(II) A 2.00-mole sample of N₂ gas at 0°C is heated to 150°C at constant pressure (1.00 atm). Determine (a) the change in internal energy, (b) the work the gas does, and (c) the heat added to it.

Key Concepts

Internal Energy

Work Done by Gas

Heat Transfer

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