Entropy Change of a Monatomic Ideal Gas Compressed Adiabatically

In summary, an expert summarizer of content provided a summary of a conversation about calculating the entropy change (ΔS) of a monatomic ideal gas that is compressed adiabatically. The conversation discussed the need to find a reversible path between the initial and final states and how the process is not quasi-static due to the constant external pressure. The expert also clarified a mistake in converting units and suggested finding a reversible path or using an equation for calculating entropy change for an ideal gas.
  • #1
Basis
3
0

Homework Statement



A monatomic ideal gas is compressed adiabatically. What is ΔS of the gas?

n = 1.00
vi = 10.00 L
vf = 4.00 L
Constant External Pressure = 50.00 atm


Homework Equations



ΔU = q + w = 0 + (-PextΔV)
ΔS = ∫[(qrev)/T]dQ


The Attempt at a Solution



In order to calculate ΔS you have connect two thermodynamic states with a reversible path. I assumed that the ΔS of the system is 0 because dQ is 0 for adiabatic processes). Then second state focuses on ΔS of the surroundings that has an external pressure associated with it.

I found Tf = 322.06 K by using the ΔU equation

I am not sure what to do at this point, because every single attempt I made has led me to a negative ΔS.
 
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  • #2
Basis said:

Homework Statement



A monatomic ideal gas is compressed adiabatically. What is ΔS of the gas?

n = 1.00
vi = 10.00 L
vf = 4.00 L
Constant External Pressure = 50.00 atm

Homework Equations



ΔU = q + w = 0 + (-PextΔV)
ΔS = ∫[(qrev)/T]dQ

The Attempt at a Solution



In order to calculate ΔS you have connect two thermodynamic states with a reversible path. I assumed that the ΔS of the system is 0 because dQ is 0 for adiabatic processes). Then second state focuses on ΔS of the surroundings that has an external pressure associated with it.

I found Tf = 322.06 K by using the ΔU equation

I am not sure what to do at this point, because every single attempt I made has led me to a negative ΔS.
You cannot assume that the ΔS = 0 because the process is not reversible. Since external pressure is constant and internal pressure varies, the compression is not quasi-static. A reversible path between the initial and final states is not adiabatic - there is heat flow.

I am not sure how you are calculating the final temperature. Are you given the initial temperature or the initial (internal) pressure of the gas?

AM
 
  • #3
Yeah I started thinking it over and it wouldn't make sense. I thought I had read somewhere that if it is reversible it is 0. I am sorry I thought I had specified that initial temperature. It is at 298 K

I used

ΔU = -PextΔV
ΔU = =-(50.00 atm)(4.00 L - 10.00 L) = 300 J
300 J = n(3/2)R(Tf - Ti)
300 J = (1.00 mole)(3/2)(8.314 J/Kmol)(Tf - 298 K)
Tf = 322.06 K

But the question is given that it is compressed adiabatically so that's where my problem lies
 
  • #4
Basis said:
Yeah I started thinking it over and it wouldn't make sense. I thought I had read somewhere that if it is reversible it is 0. I am sorry I thought I had specified that initial temperature. It is at 298 K
If the process is reversible the ΔS of the system + surroundings will be 0. As I said, this is not a reversible process. To calculate entropy change you have to first find a reversible path between the beginning and end states. So you have to find the pressure, volume and temperature of the gas initially and at the end and then find a reversible path from the initial to final states.

I used

ΔU = -PextΔV
ΔU = =-(50.00 atm)(4.00 L - 10.00 L) = 300 J
Whoa! You need MKS units to get an answer in Joules: Pressure is in Pascals and Volume is in m3. 1 atm = 101.325 x 103 Pa. and 1 L = 10-3 m3.

300 J = n(3/2)R(Tf - Ti)
300 J = (1.00 mole)(3/2)(8.314 J/Kmol)(Tf - 298 K)
Tf = 322.06 K

But the question is given that it is compressed adiabatically so that's where my problem lies
Your method is ok. You are just out by a few orders of magnitude with the units. (R = 8.314 J/mol-K). Tf is much higher than 322K.

Apart from the units, I am not sure where you are having a problem. Adiabatic processes do not have to be quasi-static. Compression in an internal combustion engine is a good example of an irreversible adiabatic compression.

Welcome to PF by the way!

AM
 
  • #5
I came to that realization not too long ago thankfully. I am glad for the clarification. I thought I was going mad for a second. Thank you so much!
 
  • #6
You still need to get the entropy change. You have all the information you need about the initial and final states. You just need to figure out a reversible path from the initial to the final state and calculate dq/T for that path, or (less preferably) you need to remember the equation for the change of entropy for an ideal gas in terms of the volume ratio and the temperature ratio. Have you seen an equation like that?

Chet
 

1. What is entropy change?

Entropy change is a measure of the disorder or randomness of a system. In thermodynamics, it is defined as the amount of energy that cannot be converted into work during a process.

2. How is entropy change calculated for a monatomic ideal gas compressed adiabatically?

The entropy change of a monatomic ideal gas compressed adiabatically can be calculated using the equation ΔS = nCvln(Tf/Ti), where n is the number of moles of gas, Cv is the molar specific heat at constant volume, and Tf and Ti are the final and initial temperatures, respectively.

3. What is an adiabatic compression?

An adiabatic compression is a process in which a system is compressed without any heat transfer to or from the surroundings. This means that the system is insulated and no energy is lost or gained as heat during the process.

4. How does entropy change for a monatomic ideal gas during an adiabatic compression?

During an adiabatic compression, the entropy of a monatomic ideal gas decreases. This is because the gas molecules are being pushed closer together, reducing their freedom of movement and increasing their orderliness.

5. Can the entropy change of a monatomic ideal gas compressed adiabatically be negative?

Yes, the entropy change of a monatomic ideal gas compressed adiabatically can be negative. Since the entropy change is calculated as a difference between the final and initial states, it can be negative if the final temperature is lower than the initial temperature.

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