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

In summary, the conversation is discussing a physics problem involving an ideal gas and the first law of thermodynamics. The problem statement gives three variables, T, V, and P, and asks for the correct answer from a list of choices. The participants discuss the process and variables involved, with one participant suggesting that the process is isobaric and option D) is the correct answer. Another participant explains the reasoning behind this suggestion and how it relates to the first law of thermodynamics.
  • #1
Vibhor
971
40

Homework Statement



?temp_hash=462af0fbcc5d83607f18e3d0c9ee01fd.png


Homework Equations

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}## .

I do not have much idea . Please help me with the problem .

Thanks
 

Attachments

  • variables.PNG
    variables.PNG
    19.6 KB · Views: 535
Last edited:
Physics news on Phys.org
  • #2
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
Simon Bridge said:
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?

I do not understand your reply . Please elaborate .

Please let me know specifically what mistake do you find in the OP ?
 
  • #4
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
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
rude man said:
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.

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:
  • #7
Vibhor said:
As far as I am concerned I am over with the problem so please feel free to give your interpretation .
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:
  • #8
collinsmark said:
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.)

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
Vibhor said:
Are you sure ?
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
Vibhor said:
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 ?
collinsmark said:
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
For a gas, if two are given the third is determined ipso facto. Unless the mass changes. Cf. "equation of state".
 
  • #11
rude man said:
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.
For an ideal gas , if internal gas is kept constant , temperature cannot vary . None of the choices make sense .
 
Last edited:
  • #12
Vibhor said:
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 ?

That was what I was hinting at, yes. :smile:

rude man said:
For a gas, if two are given the third is determined ipso facto. Unless the mass changes. Cf. "equation of state".
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:
  • #13
collinsmark said:
That was what I was hinting at, yes.
Do you think option D) is consistent with First Law of Thermodynamics ?
 
  • #14
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.)

Vibhor said:
Do you think option D) is consistent with First Law of Thermodynamics ?
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:
  • #15
Btw, I'm not claiming I'm right. I'm just saying that that is how I interpret the problem.
 

Related to Thermodynamics -- Relation between ##dT## and ##dV##

1. What is the relation between dT and dV in thermodynamics?

In thermodynamics, dT and dV are related by the equation dT = TdS/CV, where T is temperature, S is entropy, and CV is the heat capacity at constant volume. This equation shows how a change in temperature (dT) is related to a change in volume (dV) in a system at constant entropy and heat capacity.

2. How does the relation between dT and dV affect the behavior of a gas?

The relation between dT and dV is important in determining the behavior of a gas because it is directly related to the gas's internal energy. A change in temperature (dT) can result in a change in volume (dV) and vice versa, which can affect the pressure and volume of the gas according to the ideal gas law. This relation is also used in calculating changes in the internal energy of a gas during thermodynamic processes.

3. Can the relation between dT and dV be applied to all thermodynamic systems?

The relation between dT and dV is a fundamental concept in thermodynamics and can be applied to all systems, including gases, liquids, and solids. However, the specific equation may vary depending on the system and its properties, such as heat capacity and entropy.

4. How does the relation between dT and dV relate to the first law of thermodynamics?

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. The relation between dT and dV is used to calculate the change in internal energy, which is an important component of the first law. Additionally, the first law can be used to derive the relation between dT and dV for various thermodynamic processes.

5. Can the relation between dT and dV be used to predict the behavior of a system?

The relation between dT and dV is a mathematical relationship that can be used to calculate the changes in temperature and volume in a system. It can also be used to predict the behavior of a system under specific conditions, such as changes in temperature, pressure, or volume. However, other factors, such as external forces and non-ideal behavior, may also impact the behavior of a system, so it should be used in conjunction with other thermodynamic principles and equations.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
794
  • Introductory Physics Homework Help
Replies
11
Views
738
Replies
1
Views
740
  • Introductory Physics Homework Help
Replies
4
Views
681
  • Classical Physics
Replies
6
Views
842
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Thermodynamics
Replies
4
Views
1K
Back
Top