Ideal Gas & Heat Capacity Question

[SSerenity]

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The answer is D!

There are 2 points I don't understand. From a previous question, I know that the work done from point A to B is (-300x200) = -600J.

Work = PV.
So why do we multiply the AVERAGE of Pressure by the CHANGE in volume? And Not the change in pressure x change in volume?

And for the second part of my question....
The pressure-volume change by the gas can be described with PV=nRT
And we can solve for the change in temperature required to go from A to B.

Since temperature is only an indirect measurement of heat, we would also need the specific heat to find the 'heat flow' into the gas, with Q=MCdT

That formula will allow us turn our temperature from PV=nRT into Joules absorbed (q).

What confuses me is, if our Pressure-Volume Change corresponds to a certain amount of work energy, and this can be calculated by W=PV.... then shouldnt any corresponding Temperature change be proportional to your PV value? Why do we still need the specific heat capacity to find the heat flow?

Thanks!!

 

[DrBTS]

What a tricky @#)$* question! Haha.

edited with a much more simple response:

TL;DR
The 200 value is arbitrary, since you dont know the type of gas it is so you can't figure out the mass or the specific heat. Choice B is out because the system is isothermal and you're perturbing the inverse proportionality of P and V, so temp must be increased (heat must be added) to increase P as V increases, so choice A is out.

 

[SSerenity]

What a tricky @#)$* question! Haha.

What really confused me is how you can have pressure and volume being directly proportional for an ideal gas in a system because of the ideal gas equation but that helped me realize the answer. I'm assuming this problem has some independent nozzle adjustment that can increase the pressure as volume increases hypothetically simultaneously and thus measure the effect which explains negative work being done.

So:
E = q + w
So what I'm assuming here is we are messing with the concept of an "isothermal system" where the temperature (KE) of the system remains constant with increasing pressure and volume in order to offset these changes, thus either releasing/ absorbing heat to/from the environment out/in the system. Normally as you increase the volume of the container system, the pressure would decrease to maintain a constant temperature. Alternatively if you decrease the volume, the pressure would increase. It also holds that if you increase the pressure, the volume would have to decrease etc.

Here we are increasing the pressure but also INCREASING the volume, so while the heat is free to enter or leave the system, this initial paradigm of isothermal systems is perturbed, thus choice C is out. Given the ideal gas law, PV = nRT. if you increase the PV then T would increase. Also you can think about it logically with the isothermal system stated above, as you increase the temperature of the system (without changing V), more KE, more pressure increase. And here you're proportionally INCREASING the pressure (where normally volume would have to decrease) and INCREASING the volume, so more and more heat would have to be added. I hope this makes sense, but that helps you to eliminate A (question states into the gas, so your perspective is in the system). Now, you know it's a positive heat value and you could calculate with q=mc delta T, where delta T must be positive to further prove my reasoning, but you dont know what type of gas it is (can't calculate the mass via ideal gas law, and while you can calculate the temperature change that doesnt really matter because ultimately: answer choice D which is true you dont know the specific heat value of the gas and that value depends on the heat flow).

Thanks. I understand how energy must be put into the system. But from the graph, we can see that the work done from A to B is -600 Joules (gas is doing work on the environment) At the same time, the pressure is increasing, so the heat absorbed must be a positive value.

So like you said,
E = -600 + q correct?

The equation PV=nRT will give us the change in T from A to B. Does this change in temperature correspond to "q" via specific heat?

How can we use W=PV and PV=nRT to get two separate values, to be treated differently. Shouldnt they reflect the same value, because

W=PV=nRT correct?

 

[FeinMS]

Thanks. I understand how energy must be put into the system. But from the graph, we can see that the work done from A to B is -600 Joules (gas is doing work on the environment) At the same time, the pressure is increasing, so the heat absorbed must be a positive value.

So like you said,
E = -600 + q correct?

The equation PV=nRT will give us the change in T from A to B. Does this change in temperature correspond to "q" via specific heat?

How can we use W=PV and PV=nRT to get two separate values, to be treated differently. Shouldnt they reflect the same value, because

W=PV=nRT correct?

Work=integral(P*dV). Work equals P*delta V only when pressure is constant. In this case the pressure isn't constant, so Work is simply area under the curve.

Idk if my solution is correct, but I think the q can be found by:
Delta U=3/2*nR*delta T (assuming monatomic ideal gas)
Delta U + W= q
We're given n, delta T (using ideal gas law at two points), Work (area under the curve) so why can't we find the q?

Edit: my integration was wrong.

 

[DrBTS]

W = PV and ideal gas equation are just related if you wanted to know how much work you needed to do to increase pressure and volume (as in the apparatus) and find out the change in temperature - thats all. The work is likely being done TO the system to increase the volume, amidst increased pressure, thus +600.

PV work is being done to the system, changing the pressure and volume parameters of the system and thus changing the temperature. This change is temperature is where q comes into play, as another form of energy being exchanged with the environment. Make sure you understand that W and q are different forms of energy transfer, and are only really connected with temperature variable. For example PV work is a colligative value (depends on how much substance you have in moles n) and is usually an effector into the system, where as q is specific to the type of gas and varies depending on specific heat.
But...
E = w + q = PV + mCDeltaT

 

[FeinMS]

W = PV and ideal gas equation are just related if you wanted to know how much work you needed to do to increase pressure and volume (as in the apparatus) and find out the change in temperature - thats all. The work is likely being done TO the system to increase the volume, amidst increased pressure, thus +600.

PV work is being done to the system, changing the pressure and volume parameters of the system and thus changing the temperature. This change is temperature is where q comes into play, as another form of energy being exchanged with the environment. Make sure you understand that W and q are different forms of energy transfer, and are only really connected with temperature variable. For example PV work is a colligative value (depends on how much substance you have in moles n) and is usually an effector into the system, where as q is specific to the type of gas and varies depending on specific heat.
But...
E = w + q = PV + mCDeltaT

I agree with most of this except for definition of work. Let's go back to simple dW=F*dx. Force=PA, so dW=PA*dx. A*dx is dVolume giving us dW=Pdv. integrate both sides to get W=integral(PdV). Simple definition of W=P*deltaV works only when P is constant so we can take P out of the integral, when P is not constant, this definition cannot be used. Also note that Work is simply area under the curve on P versus V graph.

In this problem, P is not constant so Work cannot be just P(delta V). Even if you try to use that, you will not know which P to use (P at A or B?) because they're different. Simply find the area under the curve W=600J

 

[DrBTS]

Work = PV.
So why do we multiply the AVERAGE of Pressure by the CHANGE in volume? And Not the change in pressure x change in volume?

Yeah I would like to understand this once and for all. You are not correct on average pressure by change in V, it's average V by change in P.

So for I understand that W is the area under the PV graph. In this case it's 1/2(Vf - Vi)(Delta P) because it forms a right triangle and thats the formula for area. Thus if if Pressure was constantly proportionally varying with volume you would be able to say W = Delta P X 1/2 (Delta V).

 

[FeinMS]

You have to also consider the square under the right triangle as area under the curve. Oh I found the wrong area above, you're right that it's 600J

 

[DrBTS]

I agree with most of this except for definition of work. Let's go back to simple dW=F*dx. Force=PA, so dW=PA*dx. A*dx is dVolume giving us dW=Pdv. integrate both sides to get W=integral(PdV). Simple definition of W=P*deltaV works only when P is constant so we can take P out of the integral, when P is not constant, this definition cannot be used. Also note that Work is simply area under the curve on P versus V graph.

In this problem, P is not constant so Work cannot be just P(delta V). Even if you try to use that, you will not know which P to use (P at A or B?) because they're different. Simply find the area under the curve W=600J

Makes Perfect Sense.. thanks! Issue resolved. 👍👍👍👍👍👍👍