The specific work done on an ideal gas during an adiabatic process in a piston-cylinder system

[joejoe121]

Homework Statement

Question: The specific work done on an ideal gas during an adiabatic process in a piston-cylinder is equal to

Relevant Equations

Q=MC*Delta T
First Law of Thermodynamics: Delta E = Q -W

a) 0
b) (P2-P1)V
c) Cp(T2-T1)
d) Cv(T2-T1) < Ans

I don't believe its B because if volume is constant, there's no work. I mostly don't understand why Cv is chosen instead of Cp.

 

[Philip Koeck]

Homework Statement: Question: The specific work done on an ideal gas during an adiabatic process in a piston-cylinder is equal to
Relevant Equations: Q=MC*Delta T
First Law of Thermodynamics: Delta E = Q -W

a) 0
b) (P2-P1)V
c) Cp(T2-T1)
d) Cv(T2-T1) < Ans

I don't believe its B because if volume is constant, there's no work. I mostly don't understand why Cv is chosen instead of Cp.

What does adiabatic mean?

 

[laser1]

Homework Statement: Question: The specific work done on an ideal gas during an adiabatic process in a piston-cylinder is equal to
Relevant Equations: Q=MC*Delta T
First Law of Thermodynamics: Delta E = Q -W

a) 0
b) (P2-P1)V
c) Cp(T2-T1)
d) Cv(T2-T1) < Ans

I don't believe its B because if volume is constant, there's no work. I mostly don't understand why Cv is chosen instead of Cp.

When constant volume, Work = 0, right? So that means that $U=Q$ at constant volume, where $Q=C_V\Delta T$. This means that $U=C_V\Delta T$ always, no matter if it is at constant volume or not.

$Q=0$ for an adiabatic process, so $U=W$, and the result follows.

 

[Philip Koeck]

When constant volume, Work = 0, right? So that means that $U=Q$ at constant volume, where $Q=C_V\Delta T$. This means that $U=C_V\Delta T$ always, no matter if it is at constant volume or not.

$Q=0$ for an adiabatic process, so $U=W$, and the result follows.

I don't think one should hand out complete solutions like that, at least not in the homework help section.
It's much better to just hint at things so that the OP has a chance to think a bit more.
Just my opinion.

 

[Chestermiller]

For an ideal gas in a reversible volume change, $$dU=nC_vdT$$ The internal energy of an ideal gas is independent of volume and pressure, and depends only on temperature.

 

[joejoe121]

What does adiabatic mean?

No heat transfer

 

[joejoe121]

When constant volume, Work = 0, right? So that means that $U=Q$ at constant volume, where $Q=C_V\Delta T$. This means that $U=C_V\Delta T$ always, no matter if it is at constant volume or not.

$Q=0$ for an adiabatic process, so $U=W$, and the result follows.

I understand that when volume change is zero, there is no work done so dU = W. However, how do I know in this process in a piston cylinder, the volume is constant?
I also have one question. If the first law of thermodynamics is dU = Q - W. Why is w positive in this case?

 

[Philip Koeck]

No heat transfer

Now if you apply that to the first law you get W = - ΔU.
If you look at posts 3 and 5 now you should get the answer you need.
Take into account that U is a state function (see post 3).

Small correction: In post 3 U should be replaced by ΔU everywhere.

 

[Philip Koeck]

Homework Statement: Question: The specific work done on an ideal gas during an adiabatic process in a piston-cylinder is equal to
Relevant Equations: Q=MC*Delta T
First Law of Thermodynamics: Delta E = Q -W

a) 0
b) (P2-P1)V
c) Cp(T2-T1)
d) Cv(T2-T1) < Ans

I don't believe its B because if volume is constant, there's no work. I mostly don't understand why Cv is chosen instead of Cp.

And actually answer d) is also wrong if T2 is the final and T1 the initial temperature in the process.
The answer should be Cv(T1-T2) since W = -ΔU.

 

[Philip Koeck]

I understand that when volume change is zero, there is no work done so dU = W. However, how do I know in this process in a piston cylinder, the volume is constant?
I also have one question. If the first law of thermodynamics is dU = Q - W. Why is w positive in this case?

Please rethink what you are writing here.
1. If W = 0 then Q = ΔU.
2. Volume is not constant in an adiabatic process.
3. From the first law you can't see whether W is positive or negative since both Q and ΔU can be positive or negative.

 

[Tiffany19]

Hey, about your question, in an adiabatic process there’s no heat exchange (Q=0), so the work done on the gas corresponds to the change in internal energy, which depends on Cv (since we're dealing with changes in constant volume, not pressure). For Cp, we'd consider heat exchange at constant pressure.

Adding to that, in adiabatic processes, the change in the gas's internal energy is linked only to temperature, so we use Cv, which directly relates to this.

 

[Philip Koeck]

Hey, about your question, in an adiabatic process there’s no heat exchange (Q=0), so the work done on the gas corresponds to the change in internal energy, which depends on Cv

That's correct.

(since we're dealing with changes in constant volume, not pressure). For Cp, we'd consider heat exchange at constant pressure.

That's not correct! For an ideal gas ΔU is always given by n C_V ΔT, no matter what the process!
Note that in an adiabatic process with an ideal gas volume is not constant!!
We never use C_P for calculating the change of inner energy of an ideal gas!

Adding to that, in adiabatic processes, the change in the gas's internal energy is linked only to temperature, so we use Cv, which directly relates to this.

That's also wrong! The internal energy of an ideal gas is always proportional to T alone, no matter what the process is.

 

[laser1]

And actually answer d) is also wrong if T2 is the final and T1 the initial temperature in the process.
The answer should be Cv(T1-T2) since W = -ΔU.

No, it's correct. The $W$ in $W=-\Delta U$ represents the work done by the gas. However, the question asked for the work done on the gas.

 

[Philip Koeck]

No, it's correct. The $W$ in $W=-\Delta U$ represents the work done by the gas. However, the question asked for the work done on the gas.

I missed that one. Very tricky!