Work Required for Adiabatic Compression

In summary, the work done when compressing one mole of gas to half of its original volume adiabatically depends on the initial temperature and is given by the formula 3/2 RT [2^(gamma-1)- 1], where R is a known constant and gamma is equal to 5/3 for a monatomic ideal gas.
  • #1
chris_avfc
85
0

Homework Statement


One mole of gas compressed to half of its original volume adiabatically, what is the work done?


Homework Equations



dE= dW + dQ (the d one the w and q should have the line though it)
pV^ α= constant = k

The Attempt at a Solution



As it is adiabatic dQ = 0
Meaning dE = dW

dW = -pdV = kv^-α dv

Integrating this with the limits 1/2 V and V.

I end up with

-κ[-v^(1-α)/(2-2α)]

Then I am stuck on where to go from there?
 
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  • #2
chris_avfc said:

Homework Statement


One mole of gas compressed to half of its original volume adiabatically, what is the work done?
Are you given the temperature of the gas initially? Can we assume it is an ideal gas? If it is an ideal gas, can you determine its Cv ?

Homework Equations



dE= dW + dQ (the d one the w and q should have the line though it)
pV^ α= constant = k
Substitute nRT/V for P to put the adiabatic condition in terms of T and V.

The Attempt at a Solution



As it is adiabatic dQ = 0
Meaning dE = dW

dW = -pdV = kv^-α dv

Integrating this with the limits 1/2 V and V.
If you express the adiabatic condition in terms of T and V, it is simply a matter of finding the ratio of final T to initial T. Since W = ΔU you just have to find the change in U.

AM
 
  • #3
Andrew Mason said:
Are you given the temperature of the gas initially? Can we assume it is an ideal gas? If it is an ideal gas, can you determine its Cv ?

Substitute nRT/V for P to put the adiabatic condition in terms of T and V.

If you express the adiabatic condition in terms of T and V, it is simply a matter of finding the ratio of final T to initial T. Since W = ΔU you just have to find the change in U.

AM

No initial temperatures.
It is an ideal gas.

I was using that p for the substitution for dw=-pdv though, so now I'm confused.
 
  • #4
chris_avfc said:
No initial temperatures.
It is an ideal gas.

I was using that p for the substitution for dw=-pdv though, so now I'm confused.
If you substitute P = nRT/V in the adiabatic condition you get:

[tex]PV^\gamma = K = nRTV^{\gamma - 1}[/tex]

So [itex]TV^{\gamma -1} = \text{constant}[/itex]

To find the final T, use:

[tex]T_f/T_i = \left(\frac{V_i}{V_f}\right)^{\gamma - 1}[/tex]

so [itex]\Delta T = T_f - T_i = T_i(\left(\frac{V_i}{V_f}\right)^{\gamma - 1} - 1)[/itex]

Using W = ΔU = nCvΔT you can develop the expression for W. You have to know what the Cv is though. If it is a monatomic ideal gas, Cv = 3R/2

AM
 
  • #5
Andrew Mason said:
If you substitute P = nRT/V in the adiabatic condition you get:

[tex]PV^\gamma = K = nRTV^{\gamma - 1}[/tex]

So [itex]TV^{\gamma -1} = \text{constant}[/itex]

To find the final T, use:

[tex]T_f/T_i = \left(\frac{V_i}{V_f}\right)^{\gamma - 1}[/tex]

so [itex]\Delta T = T_f - T_i = T_i(\left(\frac{V_i}{V_f}\right)^{\gamma - 1} - 1)[/itex]

Using W = ΔU = nCvΔT you can develop the expression for W. You have to know what the Cv is though. If it is a monatomic ideal gas, Cv = 3R/2

AM

Okay, I recognise the TV equation, just never though to combine the other two constants with the one.
The heat capacity equation is also in my notes, so I get all of that.

I end up with:

3/2 RT [2^(gamma-1)- 1]

Where the T is the initial temperature.
Obviously R is a constant that the value is known of.
You aren't given any value for T or gamma, so I am struggling to see how to go any further?

Thank you very much by the way.
 
  • #6
chris_avfc said:
I end up with:

3/2 RT [2^(gamma-1)- 1]

Where the T is the initial temperature.
Obviously R is a constant that the value is known of.
You aren't given any value for T or gamma, so I am struggling to see how to go any further?
You can't. The work needed to compress a mole of gas to half its volume depends on the initial temperature. The hotter it is, the more work that is required.

PS Does the question refer to the ideal gas as monatomic? Your use of 3R/2 for Cv is for monatomic gases. This would mean that [itex]\gamma = 5/3[/itex]

AM
 
Last edited:
  • #7
Andrew Mason said:
You can't. The work needed to compress a mole of gas to half its volume depends on the initial temperature. The hotter it is, the more work that is required.

PS Does the question refer to the ideal gas as monatomic? Your use of 3R/2 for Cv is for monatomic gases. This would mean that [itex]\gamma = 5/3[/itex]

AM



Yeah a couple of hours after I replied, I was looking over my notes and I saw that value for [itex]\gamma[/itex], substituted it in and it worked out great.

Thanks for all the help mate, it has been really useful.
 

What is "Work Required for Adiabatic Compression"?

"Work Required for Adiabatic Compression" refers to the amount of energy or work that is needed to compress a gas without any heat entering or leaving the system. This process is called adiabatic compression and it results in an increase in temperature and pressure of the gas.

What factors affect the work required for adiabatic compression?

The work required for adiabatic compression is affected by the initial volume and pressure of the gas, as well as the final volume and pressure of the gas. The type of gas being compressed and the rate at which the compression is done also play a role in determining the work required.

How is the work required for adiabatic compression calculated?

The work required for adiabatic compression can be calculated using the formula W = -PΔV, where W is the work, P is the pressure, and ΔV is the change in volume. This formula assumes that the compression is done slowly and reversibly, and that the gas follows the ideal gas law.

What are some real-life applications of adiabatic compression?

Adiabatic compression is used in various industrial processes, such as air compression in refrigerators and air conditioners, as well as in the compression of gases in engines and turbines. It is also an important concept in thermodynamics and is used to study the behavior of gases in different systems.

How does adiabatic compression differ from isothermal compression?

Adiabatic compression and isothermal compression are two different processes used to compress a gas. In adiabatic compression, no heat enters or leaves the system, resulting in an increase in temperature. In isothermal compression, the temperature remains constant as heat is allowed to enter or leave the system. This results in a lower work requirement for isothermal compression compared to adiabatic compression.

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