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PV diagram/ which labeled process has no temperature?

  • Thread starter YMMMA
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  • #1
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Homework Statement


I have had a question in my quiz that I don’t remember very well, but here’s a diagram and there’s no need for numbers in axis because it’s a concept- based question with no calculations.

The question asks if an ideal gas system moved from A-B-C-A. Where is the process by which there is no temperature

Homework Equations


PV=nRT
ΔU=Q-W

Where P is pressure, V is volume, n is the number of moles, R universal gas constant, T is the temperature, U is the internal energy, Q is heat, W is work

The Attempt at a Solution



From A to B, the pressure didn’t change, so the volume increased ( the gas expands). As indicated in the rule above, as the volume increases the temperature increases. From B to C there is a decrease in pressure, and the volume is constant. An decrease in pressure decreases the temperature. From Ato C, the volume decreased and the pressure increased, and that’s only possible if the temperature is constant or no heat is added. Am I right?
 

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Answers and Replies

  • #2
mjc123
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There is no process that has "no temperature". Process CA has no temperature change.
Maybe that's what you meant, and it was just an accidental slip. But people often make mistakes in thermodynamics questions because they fail to distinguish between thermodynamic quantities (e.g. enthalpy) and changes in those quantities. Get the habit early.
 
  • #3
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There is no process that has "no temperature". Process CA has no temperature change.
Maybe that's what you meant, and it was just an accidental slip. But people often make mistakes in thermodynamics questions because they fail to distinguish between thermodynamic quantities (e.g. enthalpy) and changes in those quantities. Get the habit early.
Yes, I meant that the temperature is constant so the change in temperature is zero.
 
  • #4
ehild
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Yes, I meant that the temperature is constant so the change in temperature is zero.
The path from C to A looks like a straight line instead of an isotherm (hyperbola) , so the temperature can change during the process, but the initial and final temperature is the same.
 
  • #5
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Is there away to distinguish between isotherm and adiabatic of not stated? Because both are hyperbola right?
 
  • #6
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For example, in this file it says that i to Y volume increased so temperature increased. From Y to f pressure decreased, so temperature decreases. From i to f, or path X, it’s a hyperbola, but it is adiabatic or isotherm. Is it even necessary to know in this particular case? Because if it is isother, change in internal energy is zero; if it is adiabatic change in internal energy equals negative the work done by th gas to expand the volume. So, in either ways, path Y is greater.
 

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  • #7
ehild
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Is there away to distinguish between isotherm and adiabatic of not stated? Because both are hyperbola right?
The equation of an isotherm is PV=const, it is a hyperbola in the P-V plane. The equation of an adiabatic process is PVγ=const, it is not a hyperbola. While PV=const is a curve symmetric to the 45°line, PVγ=const is not.
If the product of the initial volume and pressure is the same as the product of the final volume and pressure, the process can be isotherm, but not adiabatic.
 
  • #8
mjc123
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It's not possible to decide whether the path X is isothermal or adiabatic in this case, but you don't need to know in order to answer the question.
Isothermal: PV = constant (hyperbola)
Adiabatic (reversible): PVγ = constant (not hyperbola, I think)
Internal energy is a state function. What does that tell you?

ehild beat me to it!
 
  • #9
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It's not possible to decide whether the path X is isothermal or adiabatic in this case, but you don't need to know in order to answer the question.
Isothermal: PV = constant (hyperbola)
Adiabatic (reversible): PVγ = constant (not hyperbola, I think)
Internal energy is a state function. What does that tell you?

ehild beat me to it!
That it is concerned about the value at each state but not the path. That means both paths started at i and ended at f, so both are equal
 
  • #10
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That it is concerned about the value at each state but not the path. That means both paths started at i and ended at f, so both are equal
Yes. Both paths have the same change in internal energy, irrespective of the details of the two paths.
 

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