Spontaneous Expansion of Gas into Evacuated Container: Thermodynamic Analysis

In summary, the conversation discusses a gas spontaneously expanding into an evacuated container and determining the signs of various thermodynamic quantities such as delta T, delta E, delta H, delta S, q, w, and delta G. It is concluded that since the gas is considered ideal, delta E is zero and the process is assumed to be isothermal. Therefore, delta H is also zero and no added heat is needed for the gas to expand.
  • #1
coookiemonste
24
0

Homework Statement


A gas spontaneously expands into an evacuated container. Indicate whether delta T, delta E, delta H, delta S, q, w, and delta G are positive, negative, or zero.


Homework Equations


delta G < 0 (spontaneous)
delta G= delta H - TdeltaS
delta E= q + w
[/B]

The Attempt at a Solution


delta T: 0? (Is it wrong to assume this is an isothermal reaction?)
delta E: 0
delta H: -
delta S: +
q: 0
w: 0
delta G: -
Am I correct? And what difference does it make that it is in an evacuated container?
 
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  • #2
It's not the best approach to assume this is an isothermal process. Can you show that's it's an isothermal process, by assuming ideality and using what you know about the change in energy?

What's your reasoning behind delta H being negative?
 
  • #3
Hmm. If the gas is considered ideal, delta E would be zero. Since the internal E depends on T, we can assume that this is an isothermal reaction?

Should delta H be + because added heat is needed to make the gas expand?
 
  • #4
coookiemonste said:
Hmm. If the gas is considered ideal, delta E would be zero. Since the internal E depends on T, we can assume that this is an isothermal reaction?

Agreed.

coookiemonste said:
Should delta H be + because added heat is needed to make the gas expand?

How is enthalpy defined?
 
  • #5
Enthalpy is equal to the internal Energy + PV. So is delta H=0?
 
  • #6
Yep.
 
  • #7
thank you!
 

Related to Spontaneous Expansion of Gas into Evacuated Container: Thermodynamic Analysis

What is the Thermodynamics Sign Problem?

The Thermodynamics Sign Problem is a phenomenon encountered in statistical mechanics and quantum field theory, where the sign of the partition function is not always positive, leading to difficulties in interpreting physical results.

Why is the sign of the partition function important?

The partition function is a fundamental quantity used to calculate the thermodynamic properties of a system. It is related to the probability of the system being in a particular state. A negative sign of the partition function indicates that the system is more likely to be in a state with lower energy, which can lead to unphysical results.

What causes the Thermodynamics Sign Problem?

The Thermodynamics Sign Problem arises from the presence of complex factors in the partition function, which can occur due to various reasons such as the presence of fermionic degrees of freedom, non-equilibrium systems, and non-trivial boundary conditions.

How do scientists deal with the Thermodynamics Sign Problem?

Several methods have been developed to address the Thermodynamics Sign Problem, such as reweighting techniques, complex Langevin simulations, and analytical continuation. These methods aim to obtain accurate results while dealing with the complex factors in the partition function.

Can the Thermodynamics Sign Problem be completely eliminated?

No, the Thermodynamics Sign Problem cannot be completely eliminated. It is an inherent issue in certain systems and must be dealt with using appropriate techniques. However, scientists continue to research and develop new methods to mitigate its effects and obtain more accurate results.

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