Thermodynamics Debacle: Ideal Gas Expansion and Work Calculation

In summary, the gas does 285J worth of work on its surroundings. The final pressure is 4,67E4 Pa. The final ratio of the gas to its initial temperature is 1.
  • #1
a_lawson_2k
39
0

Homework Statement



Quantity of SO2 occupies volume of 5E-3 m^3 at pressure of 1,1E5 Pa. The gas expands adiabatically to volume of 1E-2 m^3 doing 285J worth of work on its surroundings. Assume the gas may be treated as ideal.
A) Find final pressure of the gas if [tex]\gamma[/tex]=1,29
B) How much work does the gas do on its surroundings?
C) What is the final ratio of the gas to its initial temperature?

Homework Equations



[tex]W=\frac {{\it P_1 V_1}-{\it P_2 V_2}}{\gamma-1}[/tex]

The Attempt at a Solution



I plugged in gamma, P1, V1 and V2, and got 4,67E4 Pa. The book says 4,50E4 Pa. Not sure how they got this.

The second part puzzles me, didn't they just tell me that the gas did 285J worth of work on its surroundings? (book says 345J)

I'm not even going to touch C until I get these out of the way.
 
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  • #2
a_lawson_2k said:

Homework Statement



Quantity of SO2 occupies volume of 5E-3 m^3 at pressure of 1,1E5 Pa. The gas expands adiabatically to volume of 1E-2 m^3 doing 285J worth of work on its surroundings. Assume the gas may be treated as ideal.
A) Find final pressure of the gas if [tex]\gamma[/tex]=1,29
B) How much work does the gas do on its surroundings?
C) What is the final ratio of the gas to its initial temperature?

Homework Equations



[tex]W=\frac {{\it P_1 V_1}-{\it P_2 V_2}}{\gamma-1}[/tex]

The Attempt at a Solution



I plugged in gamma, P1, V1 and V2, and got 4,67E4 Pa. The book says 4,50E4 Pa. Not sure how they got this.

The second part puzzles me, didn't they just tell me that the gas did 285J worth of work on its surroundings? (book says 345J)

I'm not even going to touch C until I get these out of the way.

Your equation is not correct. If the adiabatic expansion is reversible (constant equilibrium during expansion), the adiabatic condition applies and W will be:

[tex]W = \frac{P_1V_1^\gamma(V_2^{1-\gamma}-V1^{1-\gamma})}{1-\gamma}[/tex]

Try that and see if you get their answer. Ignore the 285J. It is an error if the expansion is reversible. If it is not reversible, the question makes no sense.

AM
 
  • #3
How does the [tex]P_2[/tex] come into play in that case?

EDIT: used the wrong equation in this case, [tex]p_1 V_1^\gamma=p_2 V_2^\gamma[/tex] works.
 
Last edited:

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The "Thermodynamics debacle" refers to a controversial topic in the field of thermodynamics, specifically related to the Second Law of Thermodynamics.

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The Second Law of Thermodynamics is a fundamental law of physics that states that the total entropy (disorder) of a closed system will always increase over time. This law has significant implications in areas such as energy transformation and heat transfer.

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There are several interpretations of the Second Law of Thermodynamics, leading to debates among scientists about its exact meaning and implications. Some argue that the law applies only to closed systems, while others believe it applies to all systems. There is also debate about whether the law is a fundamental principle or an emergent property of statistical mechanics.

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