EK GChem questions

[wahcha]

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Alright so I can't figure this one out:

In a free adiabatic expansion, a real gas is allowed to spread to twice its original volume with no energy transfer from the surroundings. All of the following are true concerning this process EXCEPT:

A) No work is done.
B) Increased potential energy between molecules results in decreased kinetic energy and the gas cools.
C) Entropy increases
D) The gas loses heat.

The answer is D. I thought it was a throw-up between B and D. I know the system can't lose energy to its surroundings (specifically through work via heat to the system's surroundings) but why is B true? I don't understand how potential energy increases. Also, if the gas cools ...doesn't that mean the gas loses heat? The gas's heat gets turned into potential energy... ? I can't conceptualize this at all.

Also, I know it's a lot to cover, but why is dH = q when pressure is held constant? I've kinda just accepted it up til this point, ha.

And why isn't dH = q when volume is held constant? The way I rationalized it goes something like this:

deltaH = deltaE + PdeltaV
dH = (q - w) + PdV
dH = (q - 0) + 0 when volume is constant.
So I understand why E = q when volume is held constant, but why not dH?


Can someone explain via equations? I'd appreciate any help provided. Thanks in advance! 🙂

 

[Don Draper]

Alright so I can't figure this one out:

In a free adiabatic expansion, a real gas is allowed to spread to twice its original volume with no energy transfer from the surroundings. All of the following are true concerning this process EXCEPT:

A) No work is done.
B) Increased potential energy between molecules results in decreased kinetic energy and the gas cools.
C) Entropy increases
D) The gas loses heat.

The answer is D. I thought it was a throw-up between B and D. I know the system can't lose energy to its surroundings (specifically through work via heat to the system's surroundings) but why is B true? I don't understand how potential energy increases. Also, if the gas cools ...doesn't that mean the gas loses heat? The gas's heat gets turned into potential energy... ? I can't conceptualize this at all.

Also, I know it's a lot to cover, but why is dH = q when pressure is held constant? I've kinda just accepted it up til this point, ha.

And why isn't dH = q when volume is held constant? The way I rationalized it goes something like this:

deltaH = deltaE + PdeltaV
dH = (q - w) + PdV
dH = (q - 0) + 0 when volume is constant.
So I understand why E = q when volume is held constant, but why not dH?


Can someone explain via equations? I'd appreciate any help provided. Thanks in advance! 🙂

adiabatic by definition means that you can't transfer heat. It is that simple. Remember that the MCAT will have tricky questions that you think you need to do complex reasoning, but it really hinges on a simple idea (that won't be that blatant though usually).

This makes a super strong case for D, it really becomes insignificant why the other answers are wrong. You will have scenarios like this all the time. AAMC's hardest questions will have you guessing between two PROBABLE answers, but you need to focus on the basics and chose the D answer (or the obvious one).

B will always be true if it makes sense. Conservation of energy states that PE + KE is constant. The gas is cooling and therefore is obviously losing KE. The gas cooling doesn't mean it loses heat. Q is a heat transfer to the surroundings, PV = nRT, you can keep all the heat in the system and relate V and T by charles law.

To make all your internal energy/enthalpy questions simplified, here goes. At constant volume, there is no work done and the heat of reaction is exactly the internal energy. At constant pressure, there is a small amount of PV work that can be done, therefore they aren't exactly equal (off a bit), BUT the volume change is typically so small that the enthalpy is equal to the internal energy. If you understand those simple ideas, you have learned more than you need to know for the MCAT.

The enthalpy equation allows us to show that heat is equal to internal energy at constant volume. The internal energy is MORE than that, because it can include work done. If we hold the volume constant then there is no work done and they equal each other.

Let me give you advice.

I did a little under 20 FLs. I did probably 80+ chemistry passages also with some EK1001. Never did I encounter a single question that required you to explain or know this difference (internal energy vs enthalpy). That's probably 1000 chemistry questions.

Showing you all this with equations would be a waste of time. Go do more problems, forget about stuff like this. It will not help you get any questions right on the MCAT.

 

[fizzgig]

OP fwiw i googled your first question because i thought i remembered something about temp not changing (though the other poster is correct, D HAS to be right because it's the definition of adiabatic, so the other answers are moot). in case you're dying of curiosity though, the consensus from what i found is that for an IDEAL gas temp would not change in a free adiabatic expansion. the particles have zero mass so i guess getting farther from each other has no effect. for a REAL gas you have particles with mass, so increasing distance increases PE (like planets moving apart etc), so by conservation of E KE has to drop, and T is defined as average translational KE right so hence it drops.

 

[wahcha]

Awesome thanks for the help guys.

So essentially, H = q when volume or pressure is held constant, no? Since internal energy is equal to PV work + q, no work is done in both situations? Sorry for the late reply, mcats have been dominating my life 🙂 lol

 

[Rabolisk]

Awesome thanks for the help guys.

So essentially, H = q when volume or pressure is held constant, no? Since internal energy is equal to PV work + q, no work is done in both situations? Sorry for the late reply, mcats have been dominating my life 🙂 lol

Only when the volume is held constant. Work done by (or on) a gas is PdeltaV. If there is no change in volume, no work is done. However, constant pressure still means that you can do work. In fact, if both pressure and volume varied, the math would be much tougher and would require calculus. That is, you couldn't simply figure out the work done given a change in pressure and change in volume, because work is not a state function. However, you could still figure out whether the sign of work based on whether the volume expanded or contracted.

 

[wahcha]

Good stuff thanks for the help guys! Clarified a lot for me.

 

[Don Draper]

Only when the volume is held constant. Work done by (or on) a gas is PdeltaV. If there is no change in volume, no work is done. However, constant pressure still means that you can do work. In fact, if both pressure and volume varied, the math would be much tougher and would require calculus. That is, you couldn't simply figure out the work done given a change in pressure and change in volume, because work is not a state function. However, you could still figure out whether the sign of work based on whether the volume expanded or contracted.

The reasoning here is correct.

To go a step further, the point is that in chemistry, the volume change is insignificant. Lets say we are using water as a solvent, the work done by changing the volume of the water is negligibly small. Therefore we say that the heat of reaction is basically equal to the internal energy. This is why this concept (or distinguishing between the two variables) is never tested, the values are virtually the same. Now, if we want the EXACT internal energy then we hold the volume constant and we make a table in a book.

But the MCAT isn't going to test on "oh it's really 0.1% higher". The math doesn't get that exact. But I guess they could ask you it conceptually.

Again though, 80-90% of the questions on the MCAT aren't "hard" and will not have such hard choices that you can't decipher the correct answer. You need the skill of choosing the most probable answer quickly, even when another possibility works. It's called the "rule of the obvious". I made it up, which says that when you need to chose between two answers, the obvious one is right 80-90% of the time. Which means you should be able to score 12+ on the sciences.