Thermodynamics: find the change of internal energy, the work and Q

AI Thread Summary
The discussion focuses on applying the first principle of thermodynamics to a helium gas expansion problem. The key equation ΔU = W + Q is highlighted, with clarification that the work done (W) should use the final pressure P/5. To find the change in volume (ΔV), the derived formula is ΔV = [V1(5⋅T2 - T1)] / T1. It is emphasized that Q cannot simply be calculated as T2 - T1; instead, the internal energy change must be determined first, considering the heat capacity of the gas. The conversation concludes with a consensus that the expansion is not adiabatic, as the temperature increases during the process.
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Homework Statement
A recipient with a plunger/piston is filled with Helium. The gas has initial temperature T1, pressure P and occupies a volume V1. Suddenly, the pressure changes to P/5 and the Helium expands irreversibly while heating up to T2. Find the internal energy change, the work realized during the expansion and the heat exchanged during the process.
Relevant Equations
Thermodynamics equations
According to the first principle of thermodynamics: ΔU = W + Q
Also noting that: W = -P⋅ΔV (Question: This P is the initial pressure or the final?)

To find V2:

(P1⋅V1) / T1 = (P2⋅V2) / T2 → Therefore, (P⋅V1) / T1 = [(P/5)⋅V2] / T2 → (P⋅V1) / T1 = (P⋅V2) / (5⋅T2) → V2 = (5⋅T2⋅V1) / T1

So, ΔV = V2 - V1 = [(5⋅T2⋅V1) / T1] - T1 → ΔV = [V1(5⋅T2 - T1)] / T1

Now, to replace in the formula of the work W = -P⋅ΔV do I use P or P/5 ?

And also, it is OK to use Q = T2-T1 ?
 
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zvwner said:
Homework Statement:: A recipient with a plunger/piston is filled with Helium. The gas has initial temperature T1, pressure P and occupies a volume V1. Suddenly, the pressure changes to P/5 and the Helium expands irreversibly while heating up to T2. Find the internal energy change, the work realized during the expansion and the heat exchanged during the process.
Homework Equations:: Thermodynamics equations

According to the first principle of thermodynamics: ΔU = W + Q
Also noting that: W = -P⋅ΔV (Question: This P is the initial pressure or the final?)

To find V2:

(P1⋅V1) / T1 = (P2⋅V2) / T2 → Therefore, (P⋅V1) / T1 = [(P/5)⋅V2] / T2 → (P⋅V1) / T1 = (P⋅V2) / (5⋅T2) → V2 = (5⋅T2⋅V1) / T1

So, ΔV = V2 - V1 = [(5⋅T2⋅V1) / T1] - T1 → ΔV = [V1(5⋅T2 - T1)] / T1

No. ##\Delta V=\left[5\frac{T_2}{T_1}-1\right]V_1##

Now, to replace in the formula of the work W = -P⋅ΔV do I use P or P/5 ?
You use P/5. That is the pressure that the gas does work against.
And also, it is OK to use Q = T2-T1 ?
No. You need to determine the change in internal energy first. For an ideal gas, how is the internal energy change related to the temperature change (and heat capacity at constant volume)?
 
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Many thanks for your time.
So, W = -P⋅V1[(T2/T1) - (1/5)]

Now, to determine Q and the internal energy change:
I know that:
ΔU = W + Q
and
U = cnT (internal energy equals to the heat capacity by the number of moles by the temperature) At a constant volume. But none of the other values are given. So how do I follow?
 
zvwner said:
Many thanks for your time.
So, W = -P⋅V1[(T2/T1) - (1/5)]

Now, to determine Q and the internal energy change:
I know that:
ΔU = W + Q
and
U = cnT (internal energy equals to the heat capacity by the number of moles by the temperature) At a constant volume. But none of the other values are given. So how do I follow?
For a monoatomic ideal gas like Helium, what is the molar heat capacity at constant volume in terms of R?
From the ideal gas law, what is the number of moles n in terms of P, V1, R, and T1?
 
Think of the word suddenly in the exercise statement. It implies that the expansion happens really fast, which simplifies the first law of Thermodynamics.
 
JD_PM said:
Think of the word suddenly in the exercise statement. It implies that the expansion happens really fast, which simplifies the first law of Thermodynamics.
What are you driving at here?
 
Chestermiller said:
What are you driving at here?

Adiabatic expansion
 
JD_PM said:
Adiabatic expansion
This is not necessarily an adiabatic expansion. They specify the temperature change.
 
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Chestermiller said:
This is not necessarily an adiabatic expansion.

Ahh I think I see why we cannot assume adiabatic expansion here.

I had the following P-V Diagram in mind:

Captura de pantalla (867).png


But this implies that ##T_b < T_a## and the statement says:

zvwner said:
while heating up to T2

Thus we expect ##T_b > T_a## instead.
 
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JD_PM said:
Ahh I think I see why we cannot assume adiabatic expansion here.

I had the following P-V Diagram in mind:

View attachment 254938

But this implies that ##T_b < T_a## and the statement says:
Thus we expect ##T_b > T_a## instead.
Even if it were adiabatic expansion, the P-v diagram for this problem would not look the way you have drawn it.
 
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Many thanks for all the comments! I think I got it now.
👌
 
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