Calculate Entropy Change for Adiabatic Process of Argon Gas

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In summary, the entropy of an α-ideal gas changes in the following way: as discussed in lecture and in Wolfe's Notes, the entropy of an Argon gas changes due to the volume change alone (PV=nRT) and the change in entropy due to work done on a piston (W=PdV/dt).
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bullados
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Physics Homework said:
For quasistatic processes, the entropy of an α-ideal gas changes in the following way:

ΔS = nR ln(Vf/Vi) + αnR ln(Tf/Ti)

as discussed in lecture and in Wolfe's Notes. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process.

Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m^3 to 0.029 m^3 while doing work on a piston.

a) What is the change in entropy due to the volume change alone, ignoring any effects of changing internal energy?

I have tried every combination of terms that I can think of within PV=nRT to find nR. Here are the three expressions that I have come up with, none of which are correct based on the answer to this question...

nR = 100 * 0.01e-3 / 300 (the expression with kPa as the units of pressure)

nR = 100e3 * 0.01e-3 / 300 (the expression with Pa as the units of pressure)

nR = (100e3 / 1.013e5) * 0.01e-3 / 300 (the expression with atm as the units of pressure)

Help, please! I'm not sure what I'm screwing up on! Thanks to anybody who can help with this...
 
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bullados said:
I have tried every combination of terms that I can think of within PV=nRT to find nR. Here are the three expressions that I have come up with, none of which are correct based on the answer to this question...

nR = 100 * 0.01e-3 / 300 (the expression with kPa as the units of pressure)

nR = 100e3 * 0.01e-3 / 300 (the expression with Pa as the units of pressure)

nR = (100e3 / 1.013e5) * 0.01e-3 / 300 (the expression with atm as the units of pressure)

Help, please! I'm not sure what I'm screwing up on! Thanks to anybody who can help with this...
I think it will help you if you write the units with the numbers to make sure you have the result exressed in the units you need. I assume you want nR in units of Joules/K. I think I see an extraneous e-3 in the first two, and I have no idea why you tried the third one.
 
  • #3
The e-3 is the conversion factor from m^3 to liters, which is why it's included. The third is because I've had success going from Pascals to Atmospheres before with this formula, and I'm a little n00bish with using it. Lemme re-write those with units...

nR = 100 kPa * 0.01 m^3 * (1 L / 1e-3 m^3) / 300 K
nR = 100e3 Pa * 0.01 m^3 * (1 L / 1e-3 m^3) / 300 K
nR = (100e3 Pa * (1 atm / 1.013e5 pa) ) * 0.01 m^3 * (1 L / 1e-3 m^3) / 300 K

I think I just found my mistake. I feel like an idiot for that one...
 

Related to Calculate Entropy Change for Adiabatic Process of Argon Gas

1. What is entropy in thermodynamics?

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it is often referred to as the measure of the unavailability of energy to do work.

2. How is entropy change calculated for an adiabatic process?

The entropy change for an adiabatic process can be calculated using the formula ΔS = nCv ln(T2/T1), where n is the number of moles of the gas, Cv is the molar heat capacity at constant volume, T1 is the initial temperature, and T2 is the final temperature.

3. What is the significance of the adiabatic process in thermodynamics?

The adiabatic process is significant because it describes a thermodynamic process where no heat is exchanged between the system and its surroundings. This allows for the study of the change in entropy solely due to work done on the system.

4. How does the entropy change for an adiabatic process of argon gas differ from other gases?

The entropy change for an adiabatic process of argon gas is dependent on its molar heat capacity at constant volume, which is different from other gases. This value can be experimentally determined and may vary for different gases.

5. Can the entropy change for an adiabatic process be negative?

Yes, the entropy change for an adiabatic process can be negative. This indicates a decrease in disorder or randomness in the system, which is possible if work is done on the system or if the system undergoes a reversible process.

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