Confused about enthelpy change in ideal gas

[susdu]

Homework Statement

The state of 3.75mol ideal gas changed from 0.3atm and 500K to 2.2atm and 330K under constant external pressure (Pex = 2.2atm).

Calculate ΔE, work (w) and heat(Q), ΔH, ΔS.

Homework Equations

w= -∫PexdV
ΔE=1.5nRΔT=w+Q
ΔH=ΔE+Δ(PV)
ΔS=dQrev/T

The Attempt at a Solution

I recently started a thermodynamics course and this question seems entry-level but I'm still unable to grasp some fundamental concepts. Calculating the energy change, work and heat was pretty straight forward: I used the appropriate equations to find energy change and work and then used the first law to find the heat.

Problem is when I was asked about the enthalpy change, I thought "ok, this is a constant pressure process, so the enthalpy change is equal to the heat". But apparently this is not the case.

What am I missing?

 

[Chestermiller]

It's not a constant pressure process. The pressure of the surroundings is constant, but the system pressure is not constant. You already calculated ΔE, and presumably you can calculate the initial and final volumes, so you can get Δ(pV). This will give you ΔH. You can also get ΔH another way, by using nC_pΔT, and remembering that C_p=C_v+R for an ideal gas. This should, of course, give you the same answer.

 

[susdu]

So trivial, should have noticed that. Thank you.

 

[susdu]

Can you give an example of a constant pressure process where the enthalpy change equals the heat?
(I can't seem to distinguish between the given scenarios)

 

[paranormal]

Hello! It's great that you are taking a thermodynamics course and trying to understand the concepts. Let's break down the problem and try to understand it better.

Firstly, the state of the ideal gas has changed from 0.3atm and 500K to 2.2atm and 330K. This means that the temperature and pressure have both decreased. This also means that the volume of the gas has decreased, as seen from the ideal gas law (PV=nRT).

Now, let's look at the equations you have provided. The work done by the gas is given by the equation w=-∫PexdV, which is the area under the curve on a P-V graph. In this case, since the external pressure is constant, the work done is simply the change in volume multiplied by the external pressure (w=PexΔV).

Next, the change in energy (ΔE) can be calculated using the first law of thermodynamics, which states that ΔE = Q + W. In this case, since the work done is negative (gas is doing work on the surroundings), the change in energy is negative as well, indicating a decrease in energy.

Now, for the enthalpy change (ΔH), we need to consider the fact that enthalpy (H) is defined as H = E + PV. Since the pressure and volume have both decreased, the enthalpy change will not be equal to the heat (Q). This is because the heat only takes into account the change in internal energy, while enthalpy also includes the work done by the gas.

To calculate the change in entropy (ΔS), we can use the equation ΔS = dQrev/T. Since the process is reversible, we can assume that the heat transfer (Q) is equal to the change in internal energy (ΔE) and use that value to calculate ΔS.

I hope this explanation helps you understand the concept of enthalpy change in a constant pressure process. Keep practicing and asking questions to deepen your understanding of thermodynamics. Good luck!