Change of Entropy Calculation - How to do this without Final Pressure Volume?

In summary, the problem involves a mass of 1 kg of air at an initial pressure of 4.8 bar and temperature of 150 degC, enclosed in a rigid container. The air is heated until the temperature reaches 200 degC. Using the equations for change in entropy and a finite difference approximation for heat transfer, the change in entropy is calculated to be 0.08 kJ/K and the heat transferred is estimated to be approximately 35.9 kJ. The use of a rigid container means that the volume remains constant. The values for mass and initial pressure were given but were not necessary for the calculations as the problem assumes an ideal gas.
  • #1
spggodd
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Homework Statement



I am having trouble working out the change in Entropy. The question is as follows:

A mass of 1 kg of air is initially at 4.8 bar and 150 degC and it is enclosed in a rigid container. The air is heated until the temperature is 200 degC.
Calculate the change in entropy and also estimate the heat transferred by using a finite difference approximation for the equation: Q = ∫Tds.

For air, R = 0.287 kJ/kgK and Cv = 0.718 kJ/kgK

Summary:

m = 1 kg
p1 = 4.8 Bar
T1 = 150 DegC
T2 = 200 DegC

Answers: [0.08 kj/K , ~ 35.9 kJ]

Homework Equations



s2-s1 = Cv*Ln(T2/T1) + R*Ln(v2/v1)
s2-s1 = Cp*Ln(T2/T1) - R*Ln(p2/p1)

(p1*v1)/t1 = (p2*v2)/t2

The Attempt at a Solution



I calculated initial volume by using: PV=MRT and rearranging.

V1 = 89.6875x10^-6

Next I'm not sure as you don't have an final pressure to work out final velocity.
Without either I can't see how I can use the s2-s1 equations.

I am thinking this may have something to do with the "it is enclosed in a rigid container". Does this mean constant volume or something?

Thanks in advance.

Steve
 
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  • #2
spggodd said:
I am thinking this may have something to do with the "it is enclosed in a rigid container". Does this mean constant volume or something?

Thanks in advance.

Steve

Yes. It means exactly that.
 
  • #3
Thanks for confirming that, I realized I was tripping up with the temperature. I was using degrees C rather than absolute.
 
  • #4
How come it gives values for mass an initial pressure if you don't actually end up using them?

For the next part I did:

((T1*ds) + (T2*ds))/2

This gave the correct answer but I'm concerned I'm not doing it correctly as I never used the mass or pressure values
 
  • #5
spggodd said:
How come it gives values for mass an initial pressure if you don't actually end up using them?

For the next part I did:

((T1*ds) + (T2*ds))/2

This gave the correct answer but I'm concerned I'm not doing it correctly as I never used the mass or pressure values

You actually did use the mass, but apparently you don't even know you used it. If the mass were 2 kg, would your answer have been the same? Entropy is a so-called extensive property, which means it is proportional to the amount of mass.

As far as the pressure is concerned, in real life you often have information that isn't needed. You didn't need the pressure because you were able to assume an ideal gas, but, if the gas were not ideal, the internal energy would be a function not only of temperature but also of the pressure. In a way, you used the 4.8 bars to ascertain that it was valid to treat the air as an ideal gas.
 
  • #6
Ok thank you for help, I appreciate it!
 
  • #7
Ok thank you for help, I appreciate it!
 

FAQ: Change of Entropy Calculation - How to do this without Final Pressure Volume?

1. How do you calculate change of entropy without knowing the final pressure and volume?

The change of entropy can be calculated using the equation ΔS = nCln(Tf/Ti), where n is the number of moles of the system, C is the molar heat capacity, Tf is the final temperature, and Ti is the initial temperature. This equation does not require knowledge of pressure or volume.

2. What is the significance of calculating change of entropy?

The change of entropy is a measure of the disorder or randomness in a system. It is an important concept in thermodynamics and is used to predict the direction and extent of chemical reactions and physical processes.

3. Can change of entropy be negative?

Yes, change of entropy can be negative. This indicates that the system is becoming more ordered or less random. An example of this is a gas condensing into a liquid, where the molecules become more organized and have less freedom of movement.

4. How does temperature affect change of entropy?

The change of entropy is directly proportional to the temperature difference between the initial and final states. This means that as the temperature increases, the change of entropy also increases. However, the change of entropy can also be affected by changes in pressure and volume.

5. Is change of entropy a state function?

Yes, change of entropy is a state function, which means that it only depends on the initial and final states of the system and not on the path taken to reach those states. This allows for the calculation of change of entropy without knowing the specific details of a process, such as pressure and volume.

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