# Thermodynamics -- Relation between $dT$ and $dV$

1. Jul 19, 2016

### Vibhor

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

If I consider a specific case of adiabatic process , then $dQ = 0$ and $dU = -PdV$

If we differentiate with respect to T on both sides , $\frac{dU}{dT} = -P\frac{dV}{dT}$ . Now , the closest I see is option B) , but $C_V = \left ( \frac{dU}{dT} \right )_V$ not simply $\frac{dU}{dT}$ .

Thanks

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Last edited: Jul 19, 2016
2. Jul 20, 2016

### Simon Bridge

Try using your understanding instead ... for the question to make sense, which of the variables are changing?
Which answer(s) are consistent with those variables changing?

3. Jul 20, 2016

### Vibhor

Please let me know specifically what mistake do you find in the OP ?

4. Jul 20, 2016

### Simon Bridge

The specific error is in the approach to the problem. The problem is testing your understanding of the physics and not your ability to choose the correct equation.
Since the problem does not refer to the adiabatic process, or any specific process, it would be a mistake to assume adiabatic or any specific process.
But you can attempt to use the clues in the problem statement and the possible answers along with your understanding of the physics involved to work out the best answer out of the four.

But you can try guessing what equations apply if you like.

5. Jul 24, 2016

### rude man

One of the given choices is correct if and only if and only if the internal energy is kept constant (δQ = δW).
Cf. 1st law.

6. Jul 25, 2016

### Vibhor

Are you sure ?

Even though the question is ambiguous but I can surely tell you that it is in context of an ideal gas . If you have a way of interpreting the question so as to get one of the options , please explain .

As far as I am concerned I am over with the problem so please feel free to give your interpretation .

Last edited: Jul 25, 2016
7. Jul 25, 2016

### collinsmark

The way I look at the problem, you are given three variables, T, P and V. At least two of them are allowed to vary (there are two differentials given). More-so, since only two differentials are given [in the list of choices], I'd say that only two of those variables are allowed to vary. So which one is not varying? Which of the variables is held constant? What is the process called with that particular variable held constant? [Edit: And which particular heat capacity (Cv or Cp) applies when that particular variable is held constant?]

In an ideal gas, in such a process, are T and V directly proportional to each other or inversely proportional? (That might help determine if the slope of the line is positive or negative.)

Last edited: Jul 25, 2016
8. Jul 25, 2016

### Vibhor

Hi collinsmark ,

Nice to see your reply . If I am understanding you properly , you are hinting that process is isobaric and option D) is correct .

Right ?

9. Jul 25, 2016

### rude man

We have one or two high-powered thermodynamicists in this forum so chances are if I'm wrong we'll all hear about it!

10. Jul 25, 2016

### rude man

For a gas, if two are given the third is determined ipso facto. Unless the mass changes. Cf. "equation of state".

11. Jul 25, 2016

### Vibhor

For an ideal gas , if internal gas is kept constant , temperature cannot vary . None of the choices make sense .

Last edited: Jul 25, 2016
12. Jul 26, 2016

### collinsmark

That was what I was hinting at, yes.

Yes, if you know two state variables, that is enough to find the rest.

But in this case, the problem statement specified T, V, and P. Obviously, T and V are allowed to change, even if by a small, infinitesimal amount. (Edit: this is obvious since we're finding the relationship between dV and dT.)

The problem statement never mentioned enthaply or entropy at all, so I suspect that they are not relevant. Since it mentioned P, but no mention of dP, I assume that dP is 0, and thus P is what is held constant.

Last edited: Jul 26, 2016
13. Jul 26, 2016

### Vibhor

Do you think option D) is consistent with First Law of Thermodynamics ?

14. Jul 26, 2016

### collinsmark

Putting it another way, here is how I interpret the problem:

Suppose you have a system with a roughly constant temperature T, volume V, and pressure P. If you ever so slightly "wiggle" the system's volume, how does that affect the system's temperature, all else being the same (i.e., the pressure is kept the same)? (You may ignore the system's enthalpy and entropy; they may change as they will in response to your "wiggle" in the volume under constant pressure.)

What do you think? Is there any indication in the problem statement that this is a closed system? [Edit: In my interpretation, I don't see a stipulation of a closed system.] [Another edit: Do Isobaric processes in general violate the first law of thermodynamics?]

Last edited: Jul 26, 2016
15. Jul 26, 2016

### collinsmark

Btw, I'm not claiming I'm right. I'm just saying that that is how I interpret the problem.