Calculating the change in entropy of an ideal gas under compression

In summary, the problem involves finding the final pressure and temperature of a gas with an initial pressure of 140kPa, temperature of 25C or 298K, and volume of 0.14m^3. The gas follows the polytropic relationship PV^1.25 = constant and has specific heat values of c_p = 1.041kJ/kg.K and c_v = 0.743 kJ/kg.K. The gas is also assumed to not exchange heat with the environment. By using the polytropic relationship and the ideal gas law, the final temperature can be found to be 0.0222 times the initial volume. With this information, the entropy change can be calculated using the equation Δ
  • #1
clurt
28
0

Homework Statement


Initial
pressure: 140kPa
Temperature: 25C or 298K
Volume: 0.14m^3

Final
Pressure:1.4MPa or 1400kPa

It uses index compression, n=1.25. So PV^1.25 = constant.

c_p = 1.041kJ/kg.K and c_v = 0.743 kJ/kg.K

Homework Equations


¥ = c_p/c_v
ΔS=c_v*ln(Tf/Ti) + R*ln(Vf/Vi) [i think]

The Attempt at a Solution


I found ¥ to be 1.4012
PV^n = 12 therefore Vf = 0.0222

140*0.14^1.25=11.989
1400*Vf^1.25=11.989
Vf^1.25=11.989/1400
 
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  • #3
PeterDonis said:
So far your work looks OK. Where are you stuck?
hi,

So where do i use the Vf? I don't I use "ΔS=c_v*ln(Tf/Ti) + R*ln(Vf/Vi) " as part two asks for the final temperature. So part one wants me to find the entropy, but i don't know how to do this.
 
  • #4
Are there any other equations that involve the entropy? For example, for the special case (which it seems would apply to your problem) where the gas does not exchange heat with the environment?
 
  • #5
I don't see any problem with solving for the final temperature first and then substituting for the delta S. Alternately, you can solve for the temperature ratio in terms of the volume ratio by combining the polytropic relationship with the ideal gas law:
##\Delta lnT-\Delta lnV=\Delta lnP##
##n\Delta ln V=-\Delta lnP##

Solve for ##\Delta lnV## and ##\Delta lnT## in terms of ##\Delta lnP##

Chet
 
Last edited:

1. What is entropy?

Entropy is a measure of the disorder or randomness of a system. It can also be thought of as a measure of the amount of energy that is unavailable for work in a system.

2. How is entropy related to ideal gases?

The entropy of an ideal gas is directly proportional to its temperature and volume. As the temperature and volume of an ideal gas increase, so does its entropy.

3. How is the change in entropy of an ideal gas calculated?

The change in entropy of an ideal gas under compression can be calculated using the equation ΔS = nRln(V2/V1), where n is the number of moles of gas, R is the gas constant, and V1 and V2 are the initial and final volumes, respectively.

4. How does compression affect the entropy of an ideal gas?

Compression of an ideal gas decreases its volume, which in turn increases its entropy. This is because the gas molecules become more disordered and are able to occupy a larger volume, leading to an increase in the number of microstates, or possible arrangements, of the gas molecules.

5. What factors can affect the change in entropy of an ideal gas under compression?

The change in entropy of an ideal gas under compression is affected by the initial and final volumes, the number of moles of gas, and the gas constant. Additionally, any external factors such as temperature or pressure changes can also affect the change in entropy.

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