Relatively simple question regarding Thermo (entropy and heating)

For example, in a reversible process, the entropy change of the system and surroundings are equal and opposite, so the total entropy change of the universe is 0. In an irreversible process, the entropy change of the surroundings is greater than that of the system, leading to a positive total entropy change for the universe. This means that more heat is dissipated in the surroundings in an irreversible process, making qrev > qirr.
  • #1
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Problem 1:

Homework Statement


An ideal gas is heated reversibly at constant volume from T1 to T2. Calculate the entropy change
of the system, the surroundings, and the universe.

Homework Equations


dS = dq/T (reversible change)
and dq = CvdT (constant V)

The Attempt at a Solution


I know how to do the system:
dS = dq/T, so you can substitute the heat capacity for dq to get
dS = CvdT/T and integrate to get dS of the system to be Cv * ln(T2/T1)

Here is where I'm not sure...
Wouldn't the entropy of the surroundings just be the negative of that? I figure because it's a reversible process the entropy change of the universe is 0...correct? Please correct me on that. Also, is there a better expression or more intuitive solution for the entropy of the surroundings?

Problem 2:

Homework Statement


For a particular change of state, why is qrev>qirr? That is, why is heat flow greater for a reversible
change of state rather than when the same change of state is brought about irreversibly?

Homework Equations


None that I know of.

The Attempt at a Solution


Is it because in an irreversible process, heat is often dissipated away in other forms such as work, noise, etc...? Because the system is not in equilibrium for an irr process the heat flow is not always conserved. Is this correct and is there a better explanation for this problem?

Thanks!
 
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  • #2
I think you are right in both problems.
For problem 2, you might be able to add some more quantitative arguments in terms of the second law.
 

Related to Relatively simple question regarding Thermo (entropy and heating)

1. What is entropy and how does it relate to heating?

Entropy is a measure of the amount of disorder or randomness in a system. Heating a system increases its entropy because it increases the movement and randomness of the particles within the system.

2. How is entropy calculated?

Entropy is calculated using the equation S = k ln(W), where S is the entropy, k is the Boltzmann constant, and W is the number of possible microstates of the system.

3. Can entropy decrease in a closed system?

According to the second law of thermodynamics, the total entropy of a closed system will always increase or remain constant. Therefore, entropy cannot decrease in a closed system.

4. How does heating affect the entropy of a substance?

As a substance is heated, its particles gain more energy and move faster, increasing the number of possible microstates and therefore increasing the entropy of the substance.

5. What is the relationship between entropy and energy?

Entropy and energy are related through the second law of thermodynamics, which states that the total entropy of a closed system will always increase or remain constant. In other words, as energy is transferred and transformed within a system, its entropy will also change accordingly.

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