Work Required for Adiabatic Compression

[chris_avfc]

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?

 

[Andrew Mason]

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

 

[chris_avfc]

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.

 

[Andrew Mason]

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:

$$PV^\gamma = K = nRTV^{\gamma - 1}$$

So $TV^{\gamma -1} = \text{constant}$

To find the final T, use:

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

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

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

 

[chris_avfc]

If you substitute P = nRT/V in the adiabatic condition you get:

$$PV^\gamma = K = nRTV^{\gamma - 1}$$

So $TV^{\gamma -1} = \text{constant}$

To find the final T, use:

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

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

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.

 

[Andrew Mason]

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 $\gamma = 5/3$

AM

 

[chris_avfc]

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 $\gamma = 5/3$

AM

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

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