Question

a quantity of a monatomic ideal gas undergoes a process in which both its pressure and volume are doubled. what is the energy absorbed by the heat into the gas during this process?

Answer 1

pV=nRT

Answer 2

T quadrupled

Answer 3

so delta T= 3T

Answer 4

Your question is not correctly worded. Do you mean 'what is the energy absorbed as heat by the gas during this process' ?

If that is the case, a definite answer can't be given, unless you specify exactly how the process occurs - because the heat absorbed is dependent on the path taken between the initial and final states.

If you are only asking about the change in the internal energy of the gas, then that is a well defined question, independent of the nature of the process - knowing the change in temperature you can use the relationship\[U(T) = \frac{ 3 }{ 2 }nRT\]
for a monatomic ideal gas, to determine the change in U, where n is the number of moles of gas.

Answer 5

ProfBrainstorm- that is exactly how the problem is worded in my textbook. I am truly lost on how to work this. There are no numerical values. My teacher couldn't even explain it. I have tried to use the equation you suggested but got no where with it.
I do not know what to do.

Answer 6

I wouldn't worry too much about this, it's a poor question. If that really is the accurate wording, then the person who set the question probably doesn't fully understand the physics behind it.
Just calculate the change in internal energy of the gas from the guidance we've given here, and leave it at that.

Answer 7

I can't calculate anything. There are no numerical values. The answer from the solution manual is 6PV

Answer 8

Well I calculate an increase in internal energy of the gas equal to \[\Delta U=\frac{ 9 }{ 2 }PV\] where P and V are the initial pressure and volume respectively.

Answer 9

Thank you for your help. I am not sure what to do at this point.

Answer 10

I can explain a little why the heat absorbed is path dependent.

Case A
Suppose the gas first expands to twice the volume by filling a vacuum, so it does no work. Because it does no work, its internal energy doesnt change and since U is a function of T only, for an ideal gas, then T doesn't change either. Secondly, heat has to be provided to raise the pressure (and temperature) to the final values.

Case B
On the other hand, if the gas expanded reversibly and isothermally, heat would have to be provided during the expansion phase to compensate for the work done in expanding against a piston, say. Then a further quantity of heat would still be required to raise the temperature and pressure as in the first case.

Hence, different paths mean different quantities of heat supplied.

Answer 11

Look for a better worded question, or better still, a new textbook ?

Answer 12

Not sure I have control over that. I am just a student. My teacher is not approachable and does not help much when we have questions. I have reached out to everyone and every resource I have.

Answer 13

Well you just can't answer this question any better without clarification, in my opinion, it's not your fault.

Answer 14

What is the textbook, if you don't mind me asking ?

Answer 15

Serway+Vuille College Physics Tenth Edition ISBN: 9781285737027

Answer 16

Strange, I have a book by Serway, although not the same as yours, he knows his physics.

Answer 17

I honestly think my teacher needs to rethink being a teacher. Before getting this job I don't believe he has ever taught before.

Answer 18

Start with the definition of enthalpy and I assime n is constant
H = U + PV

now U = 3/2nRT for a monoatomic ideal gas and PV = nRT...so:

H = 3/2 nRT + nRT

H = 2nRT

so delta H = 2nR delta T

Now when the P and V double , the temperature goes up by a factor of 4
That means delta T = 3T

So deltaH = 2nR(3T) = 6nRT = 6 PV

Answer 19

There are lots of resources with the internet these days, hopefully this site can help too.

Answer 20

Does the solution make sense to you @Raitcheson?

Answer 21

I just looked at it. I think I get it but honestly where you pulled everything minus P=nRT is completely unknown to me.

Answer 22

Just look them up in your text:
1) Definition of enthalpy (H = U + PV)
2) Internal energy for a monoatomic ideal gas U = 3/2nRT
3) Ideal Gas Law (PV=nRT)
That's really all that is required.
If you have any questions, just ask.

Answer 23

Sorry, @surry99, the enthalpy does not equal 2nRT, it is 5nRT/2

Answer 24

Silly error ...thanks

Answer 25

why introduce the enthalpy anyway - do you think that is what the question is asking for ?

Answer 26

How would enthalpy apply when it hasn't been discussed yet?

how do you go from 3/2nRT +nRT to 2nRT?

what is T, where does it come from?

how does it become 3T?

Answer 27

@ProfBrainstorm ...I am not sure what it is asking for just playing a hunch

Answer 28

Ok, it is asking for heat so I am going to go with the first law of thermodynamics:

delta U = q + w

Now if I graph the change in state , I can calculate the work done by the gas (system on surroundings as it expands) as the area under the graph:

Now the area under the curve which represents the work done by the gas on the surroundings is:
1/2(PV) + PV = 3/2 PV...

Now we now U = 3/2nRT = 3/2 PV for a monoatomic ideal gas:

First law says:
delta U = q + w

3/2 nR delta T = q + w but w = -3/2 PV (the negative sign is because the gas does work on the surroundings)

As we established earlier, delta T = 3T so:

3/2 nR (3T) = q + -3/2 PV
9/2 nRT = q -3/2 PV
9/2 PV = q -3/2 PV
q = 6PV