Thermodynamics Debacle: Ideal Gas Expansion and Work Calculation

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The discussion revolves around the calculation of final pressure, work done, and temperature ratio for an adiabatic expansion of SO2 gas. The initial calculations yielded a final pressure of 4.67E4 Pa, while the book states it should be 4.50E4 Pa, leading to confusion. Participants noted that the work done was reported as 285J, conflicting with the book's 345J, suggesting an error in the problem setup if the expansion is reversible. A correction was proposed using the appropriate equation for reversible adiabatic expansion, indicating that the initial approach was flawed. The conversation emphasizes the importance of using the correct thermodynamic equations for accurate results.
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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 \gamma=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



W=\frac {{\it P_1 V_1}-{\it P_2 V_2}}{\gamma-1}

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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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 \gamma=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



W=\frac {{\it P_1 V_1}-{\it P_2 V_2}}{\gamma-1}

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:

W = \frac{P_1V_1^\gamma(V_2^{1-\gamma}-V1^{1-\gamma})}{1-\gamma}

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
 
How does the P_2 come into play in that case?

EDIT: used the wrong equation in this case, p_1 V_1^\gamma=p_2 V_2^\gamma works.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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