Solving Kinetic Theory Homework: Work Done By External Agent

[zorro]

Homework Statement

The figure shows the initial state of the gas in an insulated cylinder. The gas is ideal and it is given that, Cp/Cv = 1.5. The insulated piston by the help of an external agent is slowly shifted to the right such that the piston divides the cylinder in the ratio 1 : 3. Calculate the work done by the external agent.

The Attempt at a Solution

I calculated the final temperatures
T₁ = √2/√3 T and T₂ = √2T

Increase in internal energy of the entire system of gas ,
U = (n₁+n₂)Cv(T₁ + T₂ -2T)
Cv = 2R
U = 2RT(n₁ + n₂)(√2/√3 + √2 - 2)

(n₁ + n₂)RT=2PV
therefore U=4PV(√2/√3 + √2 - 2)
The answer is U=2.5PV(√2/√3 + √2 - 2)

Please help

 

[Andrew Mason]

You will have to explain how you are getting the temperatures. Show all your work, including how you get W from \Delta U

AM

 

[zorro]

Sure. I skipped that to save time.
Let V₁ and V₂ be the final volumes of the left and right parts resp.
V₁=3V₂ and V₁ + V₂ = 2V_o
Hence V₁ = 3V_o/2 and V₂ = V_o/2

As the process is adiabatic,
For the left part, TV^\gamma-1 is constant
Using this equation, T₁ = √2/√3 T
Similarly applying the same equation for the right part,
T₂ = √2T.

Work will be done by the gas at the expense of its own internal energy ( there is no heat exchange )
W = -\DeltaU
But work is done on the system hence it is -W
W = \DeltaU

 

[dacruick]

you're right its an adiabatic process so the change in internal energy is equal to the work done. So if you have the change in U of the first chamber, and the change in U of the second chamber, it should be easy for you to find the work done.

If you have all the initial temperatures and all of the final temperatures then you have done 90% of the problem. U(T) = 3/2 * nRT

 

[zorro]

you're right its an adiabatic process so the change in internal energy is equal to the work done. So if you have the change in U of the first chamber, and the change in U of the second chamber, it should be easy for you to find the work done.

If you have all the initial temperatures and all of the final temperatures then you have done 90% of the problem. U(T) = 3/2 * nRT

I already got the concept behind the question.
My problem is that my answer doesnot match with the given answer.

 

[zorro]

I calculated n₁ + n₂ as follows-
the number of moles present in the final condition in both the parts are equal to the no. of moles present in each part initially.
so n₁ = n₂ = PV/RT
n₁ + n₂ = 2PV/RT

Substituting this in the equation,
U = 2RT(n₁ + n₂)(√2/√3 + √2 - 2),
U= 2RT x 2PV/RT (√2/√3 + √2 - 2) = 4PV(√2/√3 + √2 - 2).

Is there any mistake in calculation of number of moles?

 

[Andrew Mason]

Increase in internal energy of the entire system of gas ,
U = (n₁+n₂)Cv(T₁ + T₂ -2T)

The above part of your analysis is not correct:

W = \Delta U = n_1C_v\Delta T_1 + n_2C_v\Delta T_2 = n_1C_v(T_1 -T) + n_2C_v(T_2 - T) \ne (n_1 + n_2)C_v(T_1 + T_2 - 2T)

If n1 = n2, you will be out by a factor of 2. Are they equal?

AM

 

[Andrew Mason]

I calculated n₁ + n₂ as follows-
the number of moles present in the final condition in both the parts are equal to the no. of moles present in each part initially.
so n₁ = n₂ = PV/RT
n₁ + n₂ = 2PV/RT

Substituting this in the equation,
U = 2RT(n₁ + n₂)(√2/√3 + √2 - 2),
U= 2RT x 2PV/RT (√2/√3 + √2 - 2) = 4PV(√2/√3 + √2 - 2).

Is there any mistake in calculation of number of moles?

You should use:

\Delta U = nC_v\Delta T

You should get:

W = \Delta U = nC_v\left(\frac{\sqrt{2}}{\sqrt{3}} + \sqrt{2} - 2\right) T

where n is the number of moles in one side of the cylinder. Determine what C_v is in terms of R and \gamma and use that value. AM

 

[zorro]

You should use:

\Delta U = nC_v\Delta T

You should get:

W = \Delta U = nC_v\left(\frac{\sqrt{2}}{\sqrt{3}} + \sqrt{2} - 2\right) T

where n is the number of moles in one side of the cylinder. Determine what C_v is in terms of R and \gamma and use that value.


AM

I proceed as you told -
n = PV/RT
Cv = R/\gamma-1 = R/1.5-1 = 2R

W = nCv T(√2/√3 + √2 - 2) = PV/RT x 2R x T(√2/√3 + √2 - 2) =2PV (√2/√3 + √2 - 2)

The answer given in my book is W = 2.5PV (√2/√3 + √2 - 2)

 

[Andrew Mason]

I proceed as you told -
n = PV/RT
Cv = R/\gamma-1 = R/1.5-1 = 2R

W = nCv T(√2/√3 + √2 - 2) = PV/RT x 2R x T(√2/√3 + √2 - 2) =2PV (√2/√3 + √2 - 2)

The answer given in my book is W = 2.5PV (√2/√3 + √2 - 2)

I get the same as you. So I would say that the book answer is wrong.

The answer in terms of gamma is:

W = nC_v(\Delta T_1 + \Delta T_2) = n(\frac{R}{\gamma-1})T(\frac{\sqrt{2}}{\sqrt{3}} + \sqrt{2} -2) = \frac{nR}{\gamma-1}(\frac{PV}{nR})\left(\frac{\sqrt{2}}{\sqrt{3}} + \sqrt{2} -2\right)= \frac{PV}{\gamma-1}\left(\frac{\sqrt{2}}{\sqrt{3}} + \sqrt{2} -2\right)

AM

 

[zorro]

Alright.
Thanks