(II) A 1.00-mol sample of an ideal diatomic gas, originally at...

Question

(II) A 1.00-mol sample of an ideal diatomic gas, originally at 1.00 atm and 20°C, expands adiabatically to 1.75 times its initial volume. What are the final pressure and temperature for the gas? (Assume no molecular vibration.)

Accepted Answer

Everyone in this problem, we're asked to determine the final pressure and temperature of a two mo sample of an ideal diatomic gas that initially starts at two atmospheres and 25 °C. After it undergoes an abatic expansion to twice its initial volume. We're told to assume that there is no molecular vibration. We're given four answer choices. Option A 0.76 atmospheres and negative 20 °C. Option B 0.76 atmospheres and negative 47 °C. Option C 0.81 atmospheres and negative 47 °C and option D 0.96 atmospheres and negative 53 °C. So we're gonna get started here just by writing out what we've been given. OK. We're given a lot of information in, in this short question. So let's write out what we have. First thing we're told is that we have two moles. So N is going to be equal to two. The next thing we're told is this kind of initial state of our gas. So we have a pressure of two atmospheres and a temperature of 25 °C. So our initial pressure P knot will be two atmospheres. Our initial temperature T knot will be 25 °C. We're told that this is an adiabatic expansion and we have an ideal diatomic gas. We're told that it's going to expand to twice its initial volume. And so we're gonna undergo this audio expansion. We wanna find the final pressure and the final temperature and we know that the final volume is twice the initial volume. All right. So these are the two values we are looking for. We have everything else written out. And we want to think about how we can relate the information we know to the information we're looking for. And recall that because we have an adiabatic process here, we have an adiabatic expansion. We have a couple of different equations. One relating the temperature in volume and one relating the pressure in volume. OK. So we're gonna go ahead and use those to find each piece that's missing. So the first equation we have T knot V not to the exponent gamma minus one is equal to TFVF to the exponent gamma minus one. Now, gamma, this is the ratio of heat capacities and recall that gamma is going to be equal to the heat capacity at constant pressure divided by the heat capacity at constant volume. Now, because we have a diatomic gas, we know that the heat capacity at constant volume CV will be equal to five halves are. And because we have an ideal gas we know that the heat capacity at constant pressure CP will just be CV plus R which gives us a value of seven halves, R, combining these together, the ratio of heat capacity gamma will be seven halves are divided by five halves are, is going to be equal to seven fifths or we can write this as 1.4. And so we have our gamma value, we know the initial temperature, we want to find the final temperature. Now, we don't have the volumes V not and BF, but we do have a relationship between the two of them. So we're gonna go ahead rearrange our equation. Use what we know about the relationship between the two volumes and see where that takes us. OK. So we wanna solve for the final temperature. So we are gonna go ahead and rearrange this equation for TF. So we have TF is going to be equal to tina V not to the exponent gamma minus one divided by the F to the exponent gamma minus one. Now again, we know that this final volume is two times the initial volume. So let's go ahead and substitute that in TF will be equal to T knot V, not to the exponent gamma minus one divided by two V knot to the exponent gamma minus one. Now we can combine our V not to the exponent gamma minus one and our two V nat to the exponent gamma minus one because they have the same exponent, we can write them as a single term. So we can write TF as T knot multiplied by V, not divided by two V knot all to the exponent gamma minus one. Now the V knots will divide out. We're left with TF is equal to T knott multiplied by one half to the exponent gamma minus one. So now we've eliminated the volume from our equation entirely. OK. So what it was important here is that we knew the relationship between the initial volume and the final volume. We didn't need to know the exact numerical value of them. OK. So now we have an equation that only has the initial temperature T knot and that ratio of heat capacity is gamma, which we know now for the temperature initially tina, we wanna use the value in Kelvin. OK. So we're given the value of 25 °C. We're gonna add 273 to that in order to get the Kelvin value, which is gonna be 298 Kelvin. So we're gonna go ahead and use that and we see that the final temperature TF will be 293. Kelvin multiplied by one half to the exponent 1.4 minus one. When we work this out on our calculator, we're gonna find that that final temperature TF is about 225.842 Kelvin. OK. Now the answer choices are in degrees Celsius. So we're gonna convert back, we're gonna subtract 273 now and find that that final temperature is about negative 47.158 °C. So this is part one of the problem that we were trying to find, we found the final temperature. Now, if we take a look at our answer, choices, we can see that both B and C have the correct answer. When we round to two significant digits, they both had this negative 47 °C. We can eliminate option A and D because they did not have the correct answer, we've narrowed it down to two. And now we need to move on to the pressure in order to figure out which answer is correct. So for the pressure, we're gonna use a very similar equation. Recall that because we have an antibiotic process. Again, we have PNV nat to the exponent gamma is equal to PFVF to the exponent gamma. We're gonna work on this very similarly to we did before we know that VF is equal to two VN. So we can write PNV not to the exponent gamma is equal to PF multiplied by two V. Not to the exponent gamma. We wanna isolate for PV or PF sorry. So we can divide by two V knot to the exponent gamma. We get that PF will be P knot multiplied by V, not divided by two V. Not all to the exponent gamma. And again, combining that fraction, the V knots will divide out. And we're eliminating the volume from our equation by knowing that relationship. And we can write that the final pressure PF will be equal to the initial pressure P knot multiplied by one half to the exponent gamma. Now the pressure, the final pressure in the answer choices is given in atmospheres, the initial pressure we have is in atmosphere. So we can go ahead and leave that alone. We don't need to convert. In this case like we did with our Calvin. OK. When we're doing degrees Celsius in Calvin, it's a little bit trickier because we add to get to that value instead of multiplying. So when we're, when we're multiplying terms like we did in our equation, it can get a little bit messy. So we need to do that conversion. Whereas for the pressure to go from atmospheres to pascals, we're multiplying by something. So if we multiply and then we divide, we end up with that same value no matter what. All right. So we're gonna leave it in atmospheres, we can substitute in our values. We have that, that final pressure PF is going to be equal to the initial pressure, two atmospheres multiplied by one half to the exponent gamma, 1.4 working this out on our calculator. We get that, that final pressure is about 0.75786 atmospheres. And again, we're gonna round that to two significant digits. When we do, we can see that the correct answer is going to be option B we have a pressure of 0.76 atmospheres and a temperature of negative 47 °C. Thanks everyone for watching. I hope this video helped see you in the next one.

(II) A 1.00-mol sample of an ideal diatomic gas, originally at 1.00 atm and 20°C, expands adiabatically to 1.75 times its initial volume. What are the final pressure and temperature for the gas? (Assume no molecular vibration.)

Key Concepts

Adiabatic Process

Ideal Gas Law

Diatomic Gas Properties

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