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

[zvwner]

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 ?

 

[Chestermiller]

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)?

 

[zvwner]

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?

 

[Chestermiller]

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?

 

[JD_PM]

Think of the word suddenly in the exercise statement. It implies that the expansion happens really fast, which simplifies the first law of Thermodynamics.

 

[Chestermiller]

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?

 

[JD_PM]

What are you driving at here?

Adiabatic expansion

 

[Chestermiller]

Adiabatic expansion

This is not necessarily an adiabatic expansion. They specify the temperature change.

 

[JD_PM]

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:

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

while heating up to T2

Thus we expect $T_b > T_a$ instead.

 

[Chestermiller]

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.

 

[zvwner]

Many thanks for all the comments! I think I got it now.