TPR Ideal Gas Question

[Testosterone]

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Question
Which of the following is true for a closed flask containing both 1 mole of ideal Gas X and 1 mole of real Gas Y?

a) The total energy of X is equal to the total energy of Y.
b) The average kinetic energy of X is equal to the average kinetic energy of Y
c) The total volume available to the gases is the same as the total volume of the flask
d) Gases X and Y are at different temperatures





Answer is supposed to be B however wouldn't a real gas lose some kinetic energy to intermolecular forces? TPR's explanation sucks. They state:

"Temperature is a measure of average kinetic energy. If gases X and Y are in the same flask they must be at the same temperature, eliminating choice D and making choice A correct..."

Can someone else elaborate on this to explain why the answer is B? I strongly believed it was A.

 

[aspiringdoc09]

D is wrong as stated. C is wrong because ideal gases don't have volume, so it is no way that both can have the same volume available. A is wrong because total energy indicates the gases have a combination of potential energy, kinetic energy, or work. Besides 1 mole doesn't equate to equal masses for each gas (i.e., 1 mole of argon is heavier than one mole of hydrogen gas). So in order to compare energies you would have to know the masses of each gas. Therefore A is not possible because you don't have enough information. Only B applies because average KE = 3/2RT. No mass is needed to determine the average KE for each gas because it is independent of mass. Therefore, they would have the same KE if the T is the same because R is the gas constant. Does this help?

 

[N1st]

D is wrong as stated. C is wrong because ideal gases don't have volume, so it is no way that both can have the same volume available. A is wrong because total energy indicates the gases have a combination of potential energy, kinetic energy, or work. Besides 1 mole doesn't equate to equal masses for each gas (i.e., 1 mole of argon is heavier than one mole of hydrogen gas). So in order to compare energies you would have to know the masses of each gas. Therefore A is not possible because you don't have enough information. Only B applies because average KE = 3/2RT. No mass is needed to determine the average KE for each gas because it is independent of mass. Therefore, they would have the same KE if the T is the same because R is the gas constant. Does this help?

KEavg = 3/2 RT is only true for ideal gases.

For real gases, is temperature proportional to average kinetic energy or average translational kinetic energy?

 

[Testosterone]

Thanks, that was very helpful. I understand why A is wrong now. However, I still don't understand why equal particles of real gases have the same KE as equal particles of ideal gases at the same temperature. Aren't real gases influenced by intermolecular forces? If so, wouldn't they theoretically have less kinetic energy than a ideal gas?

 

[N1st]

Thanks, that was very helpful. I understand why A is wrong now. However, I still don't understand why equal particles of real gases have the same KE as equal particles of ideal gases at the same temperature. Aren't real gases influenced by intermolecular forces? If so, wouldn't they theoretically have less kinetic energy than a real gas?

I think real gas would have more kinetic energy for the same temperature, because being at the same temperature means having the same translational kinetic energy. Real gases get extra kinetic energy from rotation/vibration which doesn't contribute to the temperature.

 

[aspiringdoc09]

KEavg = 3/2 RT is only true for ideal gases.

For real gases, is temperature proportional to average kinetic energy or average translational kinetic energy?

Correct. The KEavg does apply to ideal gases; however, temperature is a measure of KE and two gases at the same T will have similar KEs. Also, real gases, although not ideal, can actually come close to behaving ideal based on the size of the molecule. For instance, water vapor isn't an ideal gas, but if you use the Van der waals equation for real gases, it does come seemingly close to ideal behavior. Also real gases can behave like ideal gases at ideal conditions: low pressure and high temps. For ideal gases, PV/nRT approaches 1 at any pressure, but for real gases PV/nRT can only approach 1 at very low pressures (< 1atm). Therefore, PV/nRT only approaches 1 at high temps and lower pressures.

However, the question doesn't get into all that. You can use POE to eliminate choices you know to be wrong. Just keep in mind that as long as the temp is the same for real and ideal gases, then their KEavg are similiar. Is this better?

PS: Also, when real gases behave similar to ideal gases at higher temperatures, then they don't experience the attractive forces as much. Of course the question stem doesn't tell you the conditions of the experiment, so you can only infer. But I would have selected B solely based on the KEavg=3/2RT equation.