Adiabatic expansion at constant pressure.

In summary, the conversation discusses the calculation of various thermodynamic quantities such as heat (q), work (W), change in temperature (dT), change in internal energy (dU), and change in enthalpy (dH) for the expansion of 4.00mol of oxygen in a 20L vessel at 270K against a constant pressure of 600torr. The discussion also includes the difference between using Δ(PV) and Δ(nRT) in the calculation of enthalpy, and the importance of considering the pressure of the system in these calculations.
  • #1
181
1

Homework Statement



A sample of 4.00mol of oxygen is originally confined in a 20L vessel at 270K and then undergoes adiabatic expansion against a constant pressure of 600torr until the volume has tripled. Find q, W, dT, dU, dH.

Homework Equations



U=q+w
H=U+PV

The Attempt at a Solution



q=0 (Adiabatic)

w=-pexΔV

U=w

ΔT= ΔU/Cv

ΔH=0The reason I'm posting this is because I believe that all of the above is correct, but my friend claims that ΔH is not zero. First of all, ΔH=q for a constant pressure expansion, and q=0, so ΔH=0, right?

Also:

ΔH=ΔU+Δ(PV)=w+ PΔV= -PΔV+PΔV=0

I believe oxygen is considered a perfect gas in this question, because we are expected to calculate C(v) from C(p)-C(v)=nR

So, am I right, does ΔH=0?

EDIT:

So, I get different answers depending on whether I use ΔH=ΔU+Δ(PV) or ΔH=ΔU+Δ(nRT)

Shouldn't they be the same? shouldn't Δ(PV)=Δ(nRT) ?

I've been obsessing over problems like this for weeks now. Every time I figure one thing out another thing pops up that just confuses everything for me.
 
Last edited:
Physics news on Phys.org
  • #2
LogicX said:
The reason I'm posting this is because I believe that all of the above is correct, but my friend claims that ΔH is not zero. First of all, ΔH=q for a constant pressure expansion, and q=0, so ΔH=0, right?

No, ΔH = q only if the pressure of the system stay constant. Although the external pressure is constant in this problem, the pressure of the system is not.
 
  • #3
Ygggdrasil said:
No, ΔH = q only if the pressure of the system stay constant. Although the external pressure is constant in this problem, the pressure of the system is not.

Are you talking about a reversible vs. irreversible change here?

So why does Δ(PV) not equal Δ(nRT)? Is it because to use Δ(PV), the P that you use has to be constant for the whole system? I didn't see that anywhere in my textbook.

So the P in the work equation is p(external) and in the enthalpy equation it is the pressure of the system?

EDIT: Ok, I see see that my last statement is correct, and Δ(PV) does indeed equal Δ(nRT), but since P is changing Δ(PV)≠PΔV because P is the pressure of the system. So we just use nRΔT because we know all those quantities.

I'm glad I finally cleared this up.
 
Last edited:

1. What is adiabatic expansion at constant pressure?

Adiabatic expansion at constant pressure is a thermodynamic process in which a gas expands without any heat entering or leaving the system, while the external pressure remains constant.

2. How is adiabatic expansion at constant pressure different from other types of expansion?

Adiabatic expansion at constant pressure is different from other types of expansion, such as isothermal or isobaric, because it occurs without any heat exchange with the surroundings. This means that the temperature of the gas will change during the process.

3. What is the equation for calculating work done during adiabatic expansion at constant pressure?

The equation for calculating work done during adiabatic expansion at constant pressure is W = P(V2 - V1), where W is the work done, P is the external pressure, and V2 and V1 are the final and initial volumes of the gas, respectively.

4. What is the significance of adiabatic expansion at constant pressure?

Adiabatic expansion at constant pressure has important applications in thermodynamics, as it allows for the calculation of work done and changes in internal energy of a system without the need to consider heat transfer. It is also commonly seen in natural processes, such as atmospheric air expanding as it rises in elevation.

5. How does adiabatic expansion at constant pressure affect the temperature and pressure of the gas?

During adiabatic expansion at constant pressure, the temperature of the gas decreases due to the decrease in internal energy, while the pressure remains constant. This is because the energy of the gas is used to do work on the surrounding environment, rather than increase the kinetic energy of the gas particles.

Suggested for: Adiabatic expansion at constant pressure.

Replies
4
Views
1K
Replies
5
Views
1K
Replies
2
Views
1K
Replies
11
Views
2K
Replies
3
Views
614
Replies
5
Views
1K
Replies
1
Views
1K
Back
Top