Adiabatic expansion problems

In summary, the conversation involves a student asking for help with two problems regarding ideal diatomic and monatomic gases undergoing adiabatic expansion. The student provides their attempts at solving the problems and asks for clarification on where their errors may be. They also mention using a graphing calculator to assign variables for more accurate calculations. The expert suggests using a simpler approach and guides the student towards finding the correct answer for the first problem. The expert also notes that for the second problem, the exponent in the relevant equation should be ((5/3)-1) instead of (1.4-1) due to the gas being monatomic. The student expresses gratitude for the help and states that they will work on the second problem using the expert's advice
  • #1
Matt Armstrong

Homework Statement


I do not know if I am allowed to have two question in one post, so forgive me if I am breaking a rule. These two are frustrating me as I cannot see where the error in my process is.

1) A 1.00-mol sample of an ideal diatomic gas, originally at 1.00 atm and 10 ∘C, expands adiabatically to 1.75 times its initial volume. What are the final temperatures for the gas?

2) An ideal monatomic gas, consisting of 2.8 mol of volume 8.3×10−2 m3 , expands adiabatically. The initial and final temperatures are 95 ∘C and -81 ∘C. What is the final volume of the gas?

Homework Equations



P_1*V_1 = P_2*V_2

PV = nRT

PV^γ = constant

TV^(γ-1) = constant

The Attempt at a Solution



Attempt at #1[/B]: I know we needed final pressure for this; I had calculated it and was told it was correct. Though the answer was rounded, I kept my final pressure unaltered, since the calculator I am using allows me to assign single letter variables to numerical constants. The final pressure I got was 1 * (1/1.75)^1.4 =~ .457 when rounded.

I calculated the true initial volume via the constants given; I converted the pressure (1 atm) to 101325 Pascals, and the temperature from Celsius to Kelvin. I then solved for Volume, then I multiplied by 1.75 to get the final volume, which gave me approximately .0398, but I again did not round and assigned it to a letter. I used algebra to get T_f = P_f*V_f / n*R = about 221 when rounded (like the other values, I assigned it a letter for an exact number). Then I subtracted 273.15 from that number and got about -52 degrees Celsius; rounded to two sigfigs as requested by the problem. But that is still wrong; I do not know where my error is.

Attempt at #2: I felt this was simple, T_1*(V_1)^(1.4-1) = T_2*(V_2)^(1.4-1). Once again I set up my equation by changing Celsius to Kelvin, and we already had V_1, so I felt it was simple algebra.

I solved for V_2, which came out to be about .42 (4.2 * 10^-1) which I felt was reasonable given there was such a sharp decrease in temperature. Again, I was told I was wrong and I am not sure why.
 
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  • #2
For problem 1 your answer is not very far off, but you may have too much "round off" error that has accumulated in your many separate calculations. I have not looked at problem 2 yet except to note that now you are dealing with a monatomic gas.

There is a much simpler way to get the answer using just one of the equations that you listed. You do not need to find any pressures or the values of any volumes.
 
  • #3
TSny said:
For problem 1 your answer is not very far off, but you may have too much "round off" error that has accumulated in your many separate calculations. I have not looked at problem 2 yet except to note that now you are dealing with a monatomic gas.

There is a much simpler way to get the answer using just one of the equations that you listed. You do not need to find any pressures or the values of any volumes.
For problem 1, the statement says diatomic. But, I agree that the T-V relevant equation is easier to use.
 
  • #4
TSny said:
problem 1 your answer is not very far off, but you may have too much "round off" error that has accumulated in your many separate calculations.

I am afraid I do not understand. Is "round off" error an error that accumulates from rounding off the separate answers? If so, that should not be a factor. I am using Desmos Graphing Calculator so that I can assign the full answer for each separate calculation to a variable - the values I am calculating with are up to 12-13 digits.

The true values for final pressure, volume and final temperature in Kelvin are 0.456821706524, 0.0397568566494 and 221.342781882 respectively. I calculate with these and then round at the very end. While there may still be round off errors, wouldn't they be small enough to not matter?
 
  • #5
Matt Armstrong said:
I am afraid I do not understand. Is "round off" error an error that accumulates from rounding off the separate answers? If so, that should not be a factor. I am using Desmos Graphing Calculator so that I can assign the full answer for each separate calculation to a variable - the values I am calculating with are up to 12-13 digits.

The true values for final pressure, volume and final temperature in Kelvin are 0.456821706524, 0.0397568566494 and 221.342781882 respectively. I calculate with these and then round at the very end. While there may still be round off errors, wouldn't they be small enough to not matter?
Yes, if you are calculating each step with that many sig figs, then you would not need to worry about round off error.

I agree with your value for the final pressure, but not your final volume or temperature.

As a check, try to work the problem the easy way by just using the same formula as you did in the second problem.
 
  • #6
TSny said:
Yes, if you are calculating each step with that many sig figs, then you would not need to worry about round off error.

I agree with your value for the final pressure, but not your final volume or temperature.

As a check, try to work the problem the easy way by just using the same formula as you did in the second problem.

I figured it out! Thank you! I will work on the second problem; I think I understand that since it is monatomic, the exponential will not be (1.4-1) but ((5/3)-1).
 
Last edited by a moderator:

1. What is adiabatic expansion?

Adiabatic expansion is a thermodynamic process in which a gas expands without exchanging heat with its surroundings. This results in a decrease in the gas's temperature and pressure as it expands.

2. What is the formula for calculating adiabatic expansion?

The formula for calculating adiabatic expansion is P1V1γ = P2V2γ, where P is pressure, V is volume, and γ is the adiabatic index, which depends on the type of gas.

3. How is adiabatic expansion different from isothermal expansion?

Adiabatic expansion occurs without any heat exchange, while isothermal expansion occurs at a constant temperature. This means that adiabatic expansion results in a decrease in temperature, while isothermal expansion maintains a constant temperature.

4. What are some real-world applications of adiabatic expansion?

Adiabatic expansion is used in air compressors, gas turbines, and refrigeration systems. It is also important in understanding weather patterns and atmospheric phenomena, such as the formation of clouds and thunderstorms.

5. How does adiabatic expansion relate to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. Adiabatic expansion is an example of energy conversion from internal energy (temperature and pressure) to external work. The first law also applies to adiabatic expansion, as the change in internal energy is equal to the work done by the gas.

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