medscholhopeful

New Member
10+ Year Member
Apr 12, 2006
66
0
Status
Non-Student
okay i always alway get turned around on this equation U= Q - W. This is my question from my topical

3 moles of a diatomic gas undergoes an adiabatic expansion. If the internal energy of the gas decreases by 200J what is the work done by the gas.
The ans choices
-200
66.7
-66.7
+200J- this is the answer

i chose A= -200 joules because i figured sine it is adiabatic no heat is lost so the equation above is U= -W and therefore the workdone by the gas is negative. Unfortunately this is not so according to the answer can someone pls explain..thank you!
 
Jun 9, 2009
7,402
8
Status
btw U = q + w
in an adiabatic expansion, q = 0, w = - PV and U = -w = - PV
U = -200 = -w, so w = +200
 

thebillsfan

Unseasoned Veteran
10+ Year Member
5+ Year Member
Dec 22, 2008
778
0
Status
Pre-Medical
I thought U = Q-W
this is a sticky situation. it depends on how you define W. can someone enlighten us as to how the MCAT defines W? all the books say you can define it however you want, but in a question like this, you cannot define it how you want
 

RogueUnicorn

rawr.
7+ Year Member
Jul 15, 2009
9,746
1,603
Status
Resident [Any Field]
this is a sticky situation. it depends on how you define W. can someone enlighten us as to how the MCAT defines W? all the books say you can define it however you want, but in a question like this, you cannot define it how you want
if the gas DOES the work, then it's positive. if the gas RECEIVES the work, then it's negative.
 

rocuronium

10+ Year Member
May 11, 2008
311
15
Canada
Status
Resident [Any Field]
this is a sticky situation. it depends on how you define W. can someone enlighten us as to how the MCAT defines W? all the books say you can define it however you want, but in a question like this, you cannot define it how you want
It is unlikely that this question is from an AAMC test. I highly doubt that a question on the MCAT would make you distinguish between two answers based on a controversial sign convention.

Know the concept behind the law and then the sign conventions won't be an issue.
 

RogueUnicorn

rawr.
7+ Year Member
Jul 15, 2009
9,746
1,603
Status
Resident [Any Field]
It is unlikely that this question is from an AAMC test. I highly doubt that a question on the MCAT would make you distinguish between two answers based on a controversial sign convention.

Know the concept behind the law and then the sign conventions won't be an issue.
disagree, this is VERY MCAT like where they won't let you take the easy way out by getting just the number alone. the sign convention is not controversial and is entirely conceptual.
 

rocuronium

10+ Year Member
May 11, 2008
311
15
Canada
Status
Resident [Any Field]
disagree, this is VERY MCAT like where they won't let you take the easy way out by getting just the number alone. the sign convention is not controversial and is entirely conceptual.
We disagree then. As was stated earlier, it depends on how you define work (W).
 

Compass

Squishy
Moderator Emeritus
10+ Year Member
Apr 15, 2006
3,363
3
31
Status
Sign conventions have been on every MCAT I've taken, across all subjects, be it motion, velocity, electrical charge, pH, etc.

---

U= Q - W

U = -200
Q = 0
-200 = 0 - W
-200 = - W
200 = W

No sign issues here <.<
 

rocuronium

10+ Year Member
May 11, 2008
311
15
Canada
Status
Resident [Any Field]
I didn't say that sign conventions wouldn't be on the MCAT. There's more to it than just a positive or negative sign, however. If the positive and negative signs mean different things for different people then there can be some confusion.
 

RogueUnicorn

rawr.
7+ Year Member
Jul 15, 2009
9,746
1,603
Status
Resident [Any Field]
I didn't say that sign conventions wouldn't be on the MCAT. There's more to it than just a positive or negative sign, however. If the positive and negative signs mean different things for different people then there can be some confusion.
if you stay consistent it should be no problem. i still do not understand where you are getting this supposed controversy from
 
OP
M

medscholhopeful

New Member
10+ Year Member
Apr 12, 2006
66
0
Status
Non-Student
btw U = q + w
in an adiabatic expansion, q = 0, w = - PV and U = -w = - PV
U = -200 = -w, so w = +200
yeah i get the math part of it.. what i had a question on was the convention + vs - but i think i may have figured it out.. thks
 

rocuronium

10+ Year Member
May 11, 2008
311
15
Canada
Status
Resident [Any Field]
if you stay consistent it should be no problem. i still do not understand where you are getting this supposed controversy from
I completely agree.

My point is that different sign conventions exist, and that I can understand how people can be confused, especially when people are adamant that their convention is correct. That's what I mean about controversial.

As I said from the beginning: Know the concept behind the law and then the sign conventions won't be an issue.
 
OP
M

medscholhopeful

New Member
10+ Year Member
Apr 12, 2006
66
0
Status
Non-Student
this is a sticky situation. it depends on how you define W. can someone enlighten us as to how the MCAT defines W? all the books say you can define it however you want, but in a question like this, you cannot define it how you want
I agree with you cos this is where i get thrown off, i have been consistently missing these type of questions based on this convention crap but i think i may have found a way to make it make sense.

U= Q -W
I look at it from the perspective of the surroundings instead of the system.. for instance i say when the work done by a gas is positive, it could mean that the gas is increasing the energy of the surrounding..hence value for W is positive(because energy of surroundings in increasing). When the gas is being worked on, then the surrounding is losing heat to the system.. therefore it is negative.

So using this new found logic of mine the ans above become +200 because the process is adiabatic so no heat is being exchanged, so Q= 0 which gives the equation U= -W - this expression means the work is being done by the gas which translates to surroundings gaining energy in the form of wok thus W is +200J

This might be a lil weird but when i used this method of thinking i got the right answers to 4 other questions i missed so......hope it helped.
 

RogueUnicorn

rawr.
7+ Year Member
Jul 15, 2009
9,746
1,603
Status
Resident [Any Field]
I completely agree.

My point is that different sign conventions exist, and that I can understand how people can be confused, especially when people are adamant that their convention is correct. That's what I mean about controversial.

As I said from the beginning: Know the concept behind the law and then the sign conventions won't be an issue.
gotcha
 

inaccensa

10+ Year Member
7+ Year Member
Sep 5, 2008
511
1
Status
Medical Student
I agree with you cos this is where i get thrown off, i have been consistently missing these type of questions based on this convention crap but i think i may have found a way to make it make sense.

U= Q -W
I look at it from the perspective of the surroundings instead of the system.. for instance i say when the work done by a gas is positive, it could mean that the gas is increasing the energy of the surrounding..hence value for W is positive(because energy of surroundings in increasing). When the gas is being worked on, then the surrounding is losing heat to the system.. therefore it is negative.

So using this new found logic of mine the ans above become +200 because the process is adiabatic so no heat is being exchanged, so Q= 0 which gives the equation U= -W - this expression means the work is being done by the gas which translates to surroundings gaining energy in the form of wok thus W is +200J

This might be a lil weird but when i used this method of thinking i got the right answers to 4 other questions i missed so......hope it helped.
I look at this a little differently. When the work is done by the gas, W =P delta V, we are looking at the change in volume. So, when gas done the work in moving the piston, the volume of the gas is expanding and thus the final volume is greater than the intial volume and thus the change in volume is positive and the work done by the gas is positive. Here the internal energy is decreasing.

Now when the piston is forced down, the volume of the gas decreases and thus the final volume is less than the initial volume and therefore the work done is negative. Here the internal energy is increasing.
Is this correct, i never think of this interms of internal energy
 

iPodtosis

10+ Year Member
7+ Year Member
Feb 7, 2009
240
4
Status
Medical Student
ok Please correct me

U = Q -W (this W is work done BY gas right?) what I am thinking is how much heat the system has minus the work done by the gas will be the internal energy.

W= PdV (this W is also work done by gas?) so when your dV is positive, that means the gas is doing work. and plug this W back into the U=Q-W equation, the heat is subtracting out the W that gas uses to increase the volume. resulting in a lesser internal energy

Is my concept correct?