Thermodynamics Enthelpy Problem

[Mattheo]

A frictionless piston cylinder device contains a saturated liquid (refrigerant-134a). The refrigerant is heated until it reaches to some temperature. Determine the work done and heat transferred to the refrigerant during this process.

I just want to make sure that I understand this topic. I will have to use tables for this.(a)The saturated liquid is turned into super heated gas.
(b) Heat transferred is the change in the internal energy u.
(c)Work done is the change in ΔP*V where P is pressure and V is volume

h=u+PV

Are those 3 assumptions correct?

Thanks

 

[Chestermiller]

Is the pressure held constant during the process or is the volume held constant?

Chet

 

[Mattheo]

Is the pressure held constant during the process or is the volume held constant?

Chet

Pressure is constant as the piston is free to move.

 

[Chestermiller]

Pressure is constant as the piston is free to move.

Then assumptions b and c are incorrect.

Chet

 

[Mattheo]

Then assumptions b and c are incorrect.

Chet

Why not a? There is a rise in temperature. So, vaporization is process completed. Or is it because my explanation wasn't adequate?

 

[Chestermiller]

Why not a? There is a rise in temperature. So, vaporization is process completed. Or is it because my explanation wasn't adequate?

Assumption a is OK.

 

[Mattheo]

Then assumptions b and c are incorrect.

Chet

So could you tell me where I am wrong? I believe my assumptions are correct. Energy transfer by work is in terms of force(PV) and by heat is in terms of temperature (internal energy, or is it enthalpy?)

 

[Chestermiller]

So could you tell me where I am wrong? I believe my assumptions are correct. Energy transfer by work is in terms of force(PV) and by heat is in terms of temperature (internal energy, or is it enthalpy?)

It's enthalpy for this problem.

The first law applied to this problem gives:

ΔU=Q-PΔV

What does that tell you about what Q is equal to? What does that tell you about what the work is equal to?

Based on this information, how would you go about using your tables to answer the problem questions?

Chet

 

[Mattheo]

It's enthalpy for this problem.

The first law applied to this problem gives:

ΔU=Q-PΔV

What does that tell you about what Q is equal to? What does that tell you about what the work is equal to?

Based on this information, how would you go about using your tables to answer the problem questions?

Chet

I see what you mean. Thanks a lot for making me realize my mistake. But there is one more thing that I am confused: The concept of enthalpy. It is hard to define it that's why I have difficulties of using it.

I mean, if u is the internal energy, which is a function of Temperature, how can you add that up with PV, which is again a function of Temperature (PV=RT). So u and PV are not independent from each other. Moreover, increase/decrease in temperature will have the same effect on each of them. It comes up to my mind as they are two different reflections of the same thing. Then why adding two same things to each other?

 

[Chestermiller]

I see what you mean. Thanks a lot for making me realize my mistake. But there is one more thing that I am confused: The concept of enthalpy. It is hard to define it that's why I have difficulties of using it.

It isn't anything fundamental. The fundamental function is U. But, the enthalpy has been found to be very convenient function to work with in many types of problems.

I mean, if u is the internal energy, which is a function of Temperature, how can you add that up with PV, which is again a function of Temperature (PV=RT). So u and PV are not independent from each other. Moreover, increase/decrease in temperature will have the same effect on each of them.

Actually, an increase/decrease in temperature will have a different effect on each of them. For an ideal gas,

dU = C_vdT

but

dH = dU + d(PV)= C_vdT + RdT = (C_v+R)dT = C_pdT

As you get more experience solving thermo problems, you will get a better appreciation of why it is often much more convenient to work with the enthalpy.

Chet

 

[Mattheo]

It isn't anything fundamental. The fundamental function is U. But, the enthalpy has been found to be very convenient function to work with in many types of problems.


Actually, an increase/decrease in temperature will have a different effect on each of them. For an ideal gas,

dU = C_vdT

but

dH = dU + d(PV)= C_vdT + RdT = (C_v+R)dT = C_pdT

As you get more experience solving thermo problems, you will get a better appreciation of why it is often much more convenient to work with the enthalpy.

Chet

You have been really helpful. Thanks for you time.