Help a brotha out with Gen Chem? :)

[Troyvdg]

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According to Chad, W= -P(delta)V

In the attached problem, why is W= +1000J instead of -1000J?

thanks in advance!

 

[UndergradGuy7]

This is the equation for a weak acid
HA + H20 --> H3O^+ + A^-

the equation in the answer key is for the conjugate base. It even says "the salt of a weak acid."

Like HCl is the acid. The conjugate base is Cl-. Or it can be written as NaCl since NaCl dissociates (salt).
Or H3PO4 is the acid. The salt form is Na3PO4, etc.

Anyway the question says the pH, but they don't give numbers so its all concepts. A salt (conjugate base) + H2O -> OH- + (acid form)

Like F- + H2O -> OH- + HF etc...
You have OH- in the products so the pH has to be bigger than 7.
Since it is a weak base, pH can't be 14. It is 8.

 

[Troyvdg]

This is the equation for a weak acid
HA + H20 --> H3O^+ + A^-

the equation in the answer key is for the conjugate base. It even says "the salt of a weak acid."

Like HCl is the acid. The conjugate base is Cl-. Or it can be written as NaCl since NaCl dissociates (salt).
Or H3PO4 is the acid. The salt form is Na3PO4, etc.

Anyway the question says the pH, but they don't give numbers so its all concepts. A salt (conjugate base) + H2O -> OH- + (acid form)

Like F- + H2O -> OH- + HF etc...
You have OH- in the products so the pH has to be bigger than 7.
Since it is a weak base, pH can't be 14. It is 8.

wooooow. i just noticed it said salt...I must be tired 😴

Thanks for the clarification!

 

[TK2006]

According to Chad, W= -P(delta)V

In the attached problem, why is W= +1000J instead of -1000J?

thanks in advance!

d(U) = [Q] - [W]. [EQN 1]

The "minus" in the above equation is intentional. First of all, don't even think about the minus sign. Think about what internal energy and the change in internal energy would signify. If the change in internal energy is positive, then the overall energy of the system would increase, right? If the change in internal energy is negative, then the overall energy of the system would decrease, right?

Now think about the signs that would make what I just said true given the "intentional" negative sign in front of the [W].

If you put +Q in the above equation, that would make the d(U) positive (given work is negligible). Correct? +Q indicates heat is put into the system, which would make the final value of the internal energy higher than the initial value. So the change is positive. Makes sense?

If you put -Q in the above equation, that would make the d(U) negative (given work is negligible). Heat is being released by the system, hence enthalpy change is negative. Correct? If the system is giving off heat, the system is "losing" internal energy. Correct?

Now, let's talk about work. Since the negative sign in the above equation is intentional, we have to be careful with what sign we put in front of W to make d(U) term make sense.

Here is the equation again: d(U) = Q - W [EQN 1]

I will assume negligible Q to make it easier to explain, so

d(U) = -W [EQN 2]

Notice we have kept the negative sign in front of W.

Now if we want to end up with a positive d(U) meaning "adding" internal energy to the system, then we must "insert" a negative value of W into EQN 2 as shown below:

d(U) = -(-W) [EQN 3]
d(U) = +W [EQN 4]

Arriving at EQN 4, we see that our "-W" has given us an overall increase in internal energy. Therefore work must have been done "ON" the system to increase the overall energy. In conclusions, "-W" has yielded us a situation where work has been done "ON" the system.


Now, the opposite is true. If we want to end up with a negative d(U) meaning the system has "lost" energy meaning the system has expended energy in some form. How do we end up with -d(U) from EQN 2? You put in "+W" into EQN!

d(U) = -(+W) [EQN 5]
d(U) = -W [EQN 6]

As you can see in EQN 6, we ended up with a negative value for d(U) meaning the system "lost" internal energy. So the work must have been DONE BY the system. Think about. If you are "DOING" work, then you are expending energy. To make this happen, we used "+W" value. So we say "+W" is work done "BY" the system.

+Q is heat "added" to the system
*makes d(U) positive in EQN 1 meaning the system gains energy

-Q is heat "lost" from the system.
*makes d(U) negative in EQN 1 meaning the system loses energy

+W is work done "by" the system.
*makes d(U) negative in EQN 1 meaning the system loses energy

-W is work done "on" the system
*makes d(U) positive in EQN 1 meaning the system gains energy

For the problem you put up, the gas is losing energy because it is expanding. The gas's internal energy is DECREASING! So d(U) is negative! How do we make d(U) negative with W? Make W positive so that we can get what we got in EQN 6.

I hope that helps. If this is still unclear, I highly suggest you watch Khan Academy video on this http://www.youtube.com/watch?v=Xb05CaG7TsQ&feature=channel

He answers the concept of this "intentional" negative sign around 13 minutes into the video. My advice is to consider what sign in front of d(U) you want to end up with. Then everything else falls into place.

I had trouble with this one time while studying for the DAT and made sure I understood it.

Good luck!

 

[Troyvdg]

^ wow. great step by step breakdown, thank you so much!

so basically, I dont even have to pay attention to the numbers at first (i.e. the 1000 Pa)?
I should just think about it logically first, and then make the signs of my numbers conform based on that?

 

[TK2006]

^ wow. great step by step breakdown, thank you so much!

so basically, I dont even have to pay attention to the numbers at first (i.e. the 1000 Pa)?
I should just think about it logically first, and then make the signs of my numbers conform based on that?

Well in an isobaric expansion, you know that pressure is constant. The volume changed so you know that abs(W) = abs(P*d(V)). So the magnitude of W (i.e. abs(W)) is 1000J. Now, you just need to figure out the sign.

But yea. Think about logically first. 👍

 

[Troyvdg]

Well in an isobaric expansion, you know that pressure is constant. The volume changed so you know that abs(W) = abs(P*d(V)). So the magnitude of W (i.e. abs(W)) is 1000J. Now, you just need to figure out the sign.

But yea. Think about logically first. 👍

I guess I'm still confused. Following the equation, if the system is doing work on the surroundings, as in the case of this problem, would the overall W be negative?

because as the gas gains volume, the energy decreases, and since W=Pd(V)
work would be positive, but when plugged back into the original equation of

d(E)= -W, wouldn't that make work negative?


or for the other equation,

W= -Pd(V), work would come out negative, and when plugged back into the original equation of d(E) = W, work would stay negative?

 

[Troyvdg]

^ bump for an answer to this question

 

[TK2006]

I guess I'm still confused. Following the equation, if the system is doing work on the surroundings, as in the case of this problem, would the overall W be negative?

because as the gas gains volume, the energy decreases, and since W=Pd(V)
work would be positive, but when plugged back into the original equation of

d(E)= -W, wouldn't that make work negative?


or for the other equation,

W= -Pd(V), work would come out negative, and when plugged back into the original equation of d(E) = W, work would stay negative?

Woah, woah.... slow down. I think you are over thinking this. 🙂

So the system (i.e. the gas) is doing work to expand in volume; therefore, the overall energy is decreasing since the system (i.e. the gas) is losing energy to the environment. So we want to end up with EQN 6, correct?

You want to end up with "an overall negative W" so you want to "insert" a positive value for W, as we did in EQN 5 and EQN 6, correct?

Hence work is + in this case. That's about it.👍

Now, all of this sign convention is "relative." I took the Kaplan course myself and have actually encountered this problem first hand. The reason why they want a positive value for W to "insert" is because they are using

d(U) = Q - W [EQN 1]

as their "universal" equation for the first law of thermodynamics.

If we were using

d(U) = Q + W [EQN 1**]

as the "base" eqn, then our work would be "negative" here. But that's out of the scope for this discussion.

Don't over analyze this. I think you got it the first time around!

Edit: also, don't worry about W = P*d(V) too much. Just find the magnitude of W and decide on the sign later.

 

[Troyvdg]

Woah, woah.... slow down. I think you are over thinking this. 🙂

So the system (i.e. the gas) is doing work to expand in volume; therefore, the overall energy is decreasing since the system (i.e. the gas) is losing energy to the environment. So we want to end up with EQN 6, correct?

You want to end up with "an overall negative W" so you want to "insert" a positive value for W, as we did in EQN 5 and EQN 6, correct?

Hence work is + in this case. That's about it.👍

Now, all of this sign convention is "relative." I took the Kaplan course myself and have actually encountered this problem first hand. The reason why they want a positive value for W to "insert" is because they are using

d(U) = Q - W [EQN 1]

as their "universal" equation for the first law of thermodynamics.

If we were using

d(U) = Q + W [EQN 1**]

as the "base" eqn, then our work would be "negative" here. But that's out of the scope for this discussion.

Don't over analyze this. I think you got it the first time around!

Edit: also, don't worry about W = P*d(V) too much. Just find the magnitude of W and decide on the sign later.

oh ok thanks 👍 .

so for the DAT, do they use d(U) = Q + W or d(U) = Q - W?

I thought that the latter applied to physics and engineering, while the first equation applied to chemistry 😕

 

[TK2006]

oh ok thanks 👍 .

so for the DAT, do they use d(U) = Q + W or d(U) = Q - W?

I thought that the latter applied to physics and engineering, while the first equation applied to chemistry 😕

Definitely, use the latter! The latter applies to Chem and the former applies to Physics, I believe. Either way, use the latter! 👍 Good luck!

 

[DemQ]

Definitely, use the latter! The latter applies to Chem and the former applies to Physics, I believe. Either way, use the latter! 👍 Good luck!

whats the "latter" ?

 

[Troyvdg]

whats the "latter" ?

d(E)= q - w

it means the second of two mentioned of two 😀

 

[Jay20155555]

I know this is a really OLD post but it just made me amazingly confused on all of this. I even rewatched chads vid on this and I feel like he is saying the exact opposite of what you are lol....

he said he the gas is expanding the gas is doing work on the surroundings which means its going to lose energy so delta U is negative and so is work

also I tried to use [EQN 1] for this question on chads quiz and got it wrong

 

[Lnguyen7]

I know this is a really OLD post but it just made me amazingly confused on all of this. I even rewatched chads vid on this and I feel like he is saying the exact opposite of what you are lol....

he said he the gas is expanding the gas is doing work on the surroundings which means its going to lose energy so delta U is negative and so is work

also I tried to use [EQN 1] for this question on chads quiz and got it wrong

deltaE = q + w
work = -P*deltaV. If work done by the system on the surrounding, delta V > 0 (gas expansion), so work < 0. If work done by the surrounding on the system, delta V < 0 (gas compression), so work > 0.
If heat is absorbed (surrounding to system), q > 0 and if heat is released (system to surrounding), q <0.

For the quiz, delta E = 18 + (-10) = 8J

 

[Jay20155555]

if we use that rationale, the question Troyvdg had in his first post gives us a negative work, and the answer was positive.

 

[Jay20155555]

actually I think i get it. If we do it using chads way it will work out like this..... i think.....

SOOOOO chad says to use the following versions of the equations:

DeltaE = q + w, and w = -P(DeltaV)

So in the question the first guy said we know P is 1000 and DeltaV is 1 making w = -(1000)(1), BUT we know its an expanding gas which means the gas is doing work on the surroundings making it expend energy and creating a negative w SOOOOOOOOOOOOOOO we have to factor that in making the new equation -(w) = -(1000)(1) so when you solve for w it will come out to be positive 1000.

Hopefully this is the right way of thinking about it, if anyone knows if I'm wrong please let me know!!