EK 1001 Gen Chem #338 ???

[betterfuture]

Approximately what is the net work done on the gas as it goes directly from state A to state B?

Can someone explain the steps needed to solve this problem?

I understand from A to B is an expansion because the work is being done by the gas. Work should be +, right? But it is asking for work done ON the gas? Lost. Help.

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[Gurby]

I'm not 100% sure but... If work is being done by the gas, then it would be negative work. Sort of like how when a reaction is exothermic, it's giving off heat, ∆H is negative.

The system is expanding, so it's "giving off work".

As for what is the answer...

W = -∆P∆V
W = - (200)(2) = -400 J

Is that the right answer? -400J of work being done on the gas.... Or in another way of looking at it, the gas is doing +400J of work on the piston or whatever.

 

[Bleu]

Approximately what is the net work done on the gas as it goes directly from state A to state B?

View attachment 202350
Can someone explain the steps needed to solve this problem?

I understand from A to B is an expansion because the work is being done by the gas. Work should be +, right? But it is asking for work done ON the gas? Lost. Help.

If you log on to the EK site you can check the answer for almost every question.

 

[betterfuture]

I'm not 100% sure but... If work is being done by the gas, then it would be negative work. Sort of like how when a reaction is exothermic, it's giving off heat, ∆H is negative.

The system is expanding, so it's "giving off work".

As for what is the answer...

W = -∆P∆V
W = - (200)(2) = -400 J

Is that the right answer? -400J of work being done on the gas.... Or in another way of looking at it, the gas is doing +400J of work on the piston or whatever.

Just wondering. What equation do you use?

∆U=Q-W or ∆U=Q+W

And if you can, could you expand on when to use which equation and why.

 

[betterfuture]

If you log on to the EK site you can check the answer for almost every question.

Really? Do you have to purchase through their website in order to access their answers? I bought the EK 1001 books off of Amazon so I don't know how it works.

 

[Gurby]

Just wondering. What equation do you use?

∆U=Q-W or ∆U=Q+W

And if you can, could you expand on when to use which equation and why.

It should be U = Q + W, right?

Think about what this equation is saying: Internal Energy = Heat + Work

Obviously if you heat up a substance, you are increasing its energy.
Similarly if you do positive work on the system, you are increasing its internal energy. Compressing a piston is doing positive work on the gas inside, increasing its energy because now all the molecules are closer together and colliding more, etc.

Really? Do you have to purchase through their website in order to access their answers? I bought the EK 1001 books off of Amazon so I don't know how it works.

Have you really been doing practice questions without access to the answers?

 

[betterfuture]

Haha. No, I meant to say I have the physical copies of the EK 1001 books but the answers in the back of the book are not helpful. Well, for some of the questions that is.

I have seen Internal Energy written as both ways.

So say a gas in a piston expands. This means the volume increases and the temperature decreases, correct? The Internal Energy of the system will decrease if work is being done by the piston because it's using the energy to do work instead of gaining it? But on the other hand I learned that Temperature and Volume are directly proportional, according to Charles Law. If Temp goes up, Volume goes up. Could you explain.

 

[Gurby]

Haha. No, I meant to say I have the physical copies of the EK 1001 books but the answers in the back of the book are not helpful. Well, for some of the questions that is.

So you know the answer to this question. Was I correct? I don't want to go spouting off an explanation if I'm wrong, which I very well might be.

I have seen Internal Energy written as both ways.

So say a gas in a piston expands. This means the volume increases and the temperature decreases, correct?

Instead of thinking about Charles' Law, think about the ideal gas law which takes Charles' Law into account: PV = nRT

V is only proportional to T if all the other variables stay constant.
If V is increasing, and P is remaining constant.... Then either n or T must be increasing.

The Internal Energy of the system will decrease if work is being done by the piston because it's using the energy to do work instead of gaining it?

Yes.

But on the other hand I learned that Temperature and Volume are directly proportional, according to Charles Law. If Temp goes up, Volume goes up. Could you explain.

If Temp goes up, and n (moles of gas) stays the same, then there are 2 possibilities: either Volume goes up, or Pressure goes up (or both to some extent).

 

[betterfuture]

Your answer was not correct. I wanted to see the approach. The answer in the back is weird.

The volume is increased, so work is done on the surroundings. That's negative work done on the gas. You can not use PVfinal minus PVinitial. The squares beneath the path represent the work done. There are 24 squares. Each square is worth 25 J. 24 x 25 = 600.
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What? That makes no sense. I don't want to be spoon fed everything but could you care to explain the above?

 

[Gurby]

Your answer was not correct. I wanted to see the approach. The answer in the back is weird.

The volume is increased, so work is done on the surroundings. That's negative work done on the gas. You can not use PVfinal minus PVinitial. The squares beneath the path represent the work done. There are 24 squares. Each square is worth 25 J. 24 x 25 = 600.
​

What? That makes no sense. I don't want to be spoon fed everything but could you care to explain the above?

Ahhhhhhh I see.

My formula above where I said W = ∆P∆V is wrong.

It's actually W = P∆V. It only works for a gas expanding against constant external pressure. I'm so used to plugging and chugging questions like that, I didn't even think about it.

In the graph here, we can see that both pressure and volume are increasing, so that formula doesn't apply. I don't completely understand the "area under the path" part. Smells like calculus.

 

[betterfuture]

I see.

So the W=P∆V only works under isobaric conditions? So how do I approach this question? Counting squares? And I still don't get how to find if work done by the gas is + or - and with work done on the gas? Is there a special way to approach these problems. I am beyond confused with the sign conventions.

 

[Gurby]

I think the main things to take away from this problem are:

1) Understanding why work should be negative here.
2) Understanding and recognizing that you can't use W = -P∆V here because pressure is not constant.

Just by knowing those two things, you can probably get this question right on a multiple choice test via process of elimination...

They are saying each square is worth 25J because each square represents 0.5 m^3 * 50 Pa = 25J. But I don't really understand why we count up ALL of the squares under the curve, and not just those inside the ABF triangle. This means it takes more work to go from (1, 200) -> (3, 400) than it would take to go from (1, 0) -> (3, 200). Seems weird to me.

 

[Gurby]

I see.

So the W=P∆V only works under isobaric conditions? So how do I approach this question? Counting squares? And I still don't get how to find if work done by the gas is + or - and with work done on the gas? Is there a special way to approach these problems. I am beyond confused with the sign conventions.

I don't really understand this, but I plan to just keep in mind that work = area under the curve in a Pressure vs Volume graph. It kind of makes sense because P*V gives you units of Joules. This method of finding area under the curve works for any situation, not just isobaric conditions. I guess the W = PV equation is really just a handy shortcut when certain circumstances exist.

So yes, you have to count the squares here. Or you could take the integral...

RE sign conventions... The gas is doing positive work on the environment by expanding. Thus, the environment is doing negative work on the gas. If you still don't get it, the Khan Academy videos on this look really good. I'm going to watch them to brush up: https://www.khanacademy.org/test-pr...v-diagrams-part-1-work-and-isobaric-processes

 

[aldol16]

In the graph here, we can see that both pressure and volume are increasing, so that formula doesn't apply. I don't completely understand the "area under the path" part. Smells like calculus.

The integral of a P-V graph gives work. You could just calculate the area since it's not a complex function. But generally, the reason that is is because dW = P*dV and since the variables are independent, you can separate and integrate. That gives you W = integral of P*dV, which is the area under the curve.

 

[betterfuture]

Again, if pressure is not constant, count squares. And for the internal energy equation, it's best to use ∆U = Q + W.

Any value they give you, one has to look at if it's work done by or work done on the gas. Work done by gas is negative because negative dictates losing energy. Work done by gas also means it is expanding, so it is taking away from internal energy by decreasing the temperature. Work done on gas means is positive because positive means adding energy. Work done on gas means compression, so it is adding towards internal energy by increasing the temperature (gas molecules collide at a greater degree when volume becomes smaller, pushing them together).

 

[Gurby]

The integral of a P-V graph gives work. You could just calculate the area since it's not a complex function. But generally, the reason that is is because dW = P*dV and since the variables are independent, you can separate and integrate. That gives you W = integral of P*dV, which is the area under the curve.

I see why the answer key just leaves it at "count the boxes!"

Thank you for helping out us plebs!

 

[Bleu]

Really? Do you have to purchase through their website in order to access their answers? I bought the EK 1001 books off of Amazon so I don't know how it works.

No go to forums and scroll down to the 1001 books. You have to register to post not sure about reading but it's free.

 

[betterfuture]

No go to forums and scroll down to the 1001 books. You have to register to post not sure about reading but it's free.

Thanks!

 

[RockMcat520]

Thanks!

This is just the area under the curve. find the area of the triangle plus the area of the square.
Area of triangle from A to B= 200
Area of the Square= 400
Total work is 600.

 

[betterfuture]

This is just the area under the curve. find the area of the triangle plus the area of the square.
Area of triangle from A to B= 200
Area of the Square= 400
Total work is 600.

Thanks man! I'm just going to count squares from now on since area is going to give Work and just give it a - or + sign based on what they say about the fluid's work.