1st law of thermodynamics to explain the change in specific latent heat

[songoku]

Homework Statement

Use the first law of thermodynamics to suggest, with a reason, how the specific latent heat of vaporisation of water at a pressure greater than atmospheric pressure compares with its value at atmospheric pressure

Relevant Equations

ΔU = Q + W

W = p.ΔV

Q = m.L

ΔU = ΔKE + ΔPE

ΔU = Q + W
Q = ΔU - W

I am not sure how to analyze the changes in ΔU and W. The value of p increases, but how do we know about ΔV?

For ΔU, the KE is constant and the PE increases since the molecules' separation is more but how do we compare the increase to the case of normal atmospheric pressure?

Thanks

 

[kuruman]

I am not sure how to analyze the changes in ΔU and W. The value of p increases, but how do we know about ΔV?

You need to compare the heat added when the pressure is atmospheric with the heat added when the pressure is higher than atmospheric for the same $\Delta V.$ Do you see why?

 

[songoku]

Do you see why?

I think so. By taking ΔV to the same, we can say that the work done by the system will be higher.

You need to compare the heat added when the pressure is atmospheric with the heat added when the pressure is higher than atmospheric for the same $\Delta V.$

Let me try:
For same ΔV, work done by system is higher so the work done on system will be more negative.

When the pressure is higher, the boiling point of the water increases so the particles will have more KE when boils. The PE will be the same as the case of atmospheric pressure since ΔV the same so the separation of molecules is also the same for both cases (I am not sure about the PE one).

Q = ΔU - W --> at higher pressure, ΔU increases and W will be more negative making Q has higher value so the specific latent heat of vaporization is also bigger.

Is my reasoning correct? Thanks

 

[kuruman]

I think so. By taking ΔV to the same, we can say that the work done by the system will be higher.

Why can we say that?

... and W will be more negative ...

You have defined W as the work done by the system. When the gas expands is W positive or negative?

Is my reasoning correct?

Your reasoning is muddled. In its conventional form, the First Law says "The change in energy that is in the gas (ΔU) is equal to the heat (Q) that goes in the gas minus the work (W) that is done by the gas on the environment."

When the steam expands against atmospheric pressure, $$\Delta U=Q-p_{\text{atm}}\Delta V.$$ When the steam expands by the same $\Delta V$ against pressure higher than atmospheric by an amount $p_0$, $$\Delta U'=Q'-(p_{\text{atm}}+p_0)\Delta V.$$The question is, in order to have a meaningful comparison, how do the primed quantities in the second equation compare with their unprimed counterparts in the first equation?

 

[256bits]

You have defined W as the work done by the system. When the gas expands is W positive or negative?

A brief description of convention, and why it screws up learning the stuff.

That, ΔU = Q + W, should be correct.
The two ΔU = Q - W, or ΔU = Q + W are not so incompatible as they first appear.


A definition could also be stated that any energy, added or removed, such as heat or work, to a system will change the system's energy, of which ΔU is a part.
ie ΔE = ΔU + ΔKE + ΔPE + all the other body forces acting on the system( gravity, electromagnetic, ... )

For ΔU = Q + W, if we have a cylinder lying on its side on the positive x-axis with the piston to the right at a positive location, a compression results in a force F directed in the -x direction, with dx moving in the -x direction. W = ( -F)(-dx) giving a positive value of work. An expansion would be W = (-F)(+dx) giving a negative value of work.

On the other hand, for ΔU = Q - W, just do two negative sign additions and we get ΔU = Q - (- W). We then have a definition of work as positive going out of the system ( expansion W = ( -)( -F)(+dx) and negative if going in (compression W = ( -)( -F)(-dx). As per the definition of the First Law for thermodynamics you have stated normally used in most cases has the ΔU = Q - W form.

Engineers like ΔU = Q - W, for the following reason: Determining how much work output W can be produced by spending money on heat input Q, giving a more reasonable way to work with the values, and determination if the machine is worth building for profitability.

Chemists like the first formula ΔU = Q + W. Some kind of enthalpy of formation of chemical reactions.

It beffudled me to no end with the switch from chemistry with ΔU = Q + W, to ΔU = Q - W in engineering, and still does at times, wondering at times is the system doing the work or the environment.

 

[anuttarasammyak]

You need to compare the heat added when the pressure is atmospheric with the heat added when the pressure is higher than atmospheric for the same ΔV. Do you see why?

Since latent heat is expressed in J/kg, I think we need to convert the amount into mass.

 

[kuruman]

It beffudled me to no end with the switch from chemistry with ΔU = Q + W, to ΔU = Q - W in engineering ...

It's not that befuddling if you think of $\Delta U = Q-W$ as saying that "the Joules that end up inside are equal to the Joules that go in as heat less the Joules that come out as work." This specifies the sign convention that positive $Q$ means "Joules go in" and positive $W$ means "Joules come out or the gas does work on the environment".

It's the check balancing principle, what stays in the account is what you put in less what you take out.

 

[songoku]

Why can we say that?

Because work done by the system = p . ΔV so higher p means higher W

You have defined W as the work done by the system. When the gas expands is W positive or negative?

I wrote the formula as ΔU = Q + W so shouldn't the W there is work done on the system?

When the steam expands against atmospheric pressure, $$\Delta U=Q-p_{\text{atm}}\Delta V.$$ When the steam expands by the same $\Delta V$ against pressure higher than atmospheric by an amount $p_0$, $$\Delta U'=Q'-(p_{\text{atm}}+p_0)\Delta V.$$The question is, in order to have a meaningful comparison, how do the primed quantities in the second equation compare with their unprimed counterparts in the first equation?

My idea is to find comparison of Q and Q' through the comparison of ΔU and ΔU' and what I can think of is using ΔU = ΔKE + ΔPE

ΔU' = ΔKE' + ΔPE' --> Since ΔV and ΔV' are the same, I think ΔPE and ΔPE' are also the same. ΔKE' > ΔKE due to higher boiling point so ΔU' > ΔU

By taking W as the work done on the system, Q = ΔU - W
W' < W and ΔU' > ΔU so Q' > Q and then L' > L

 

[kuruman]

Since latent heat is expressed in J/kg, I think we need to convert the amount into mass.

I don't think that is necessary. To make sure that the comparison between the two cases is meaningful, you have to assume that the same mass of water at 100 ^oC is converted to steam at 100 ^oC.

 

[kuruman]

Because work done by the system = p . ΔV so higher p means higher W

Higher than what? You have to put it together with specific equations, not words.

I wrote the formula as ΔU = Q + W so shouldn't the W there is work done on the system?

Yes, my oversight.

My idea is to find comparison of Q and Q' through the comparison of ΔU and ΔU' and what I can think of is using ΔU = ΔKE + ΔPE

What exactly are the meanings of ΔU, ΔKE and ΔPE when you convert an amount of water into steam? You need to be clear about that before manipulating equations. Post #9 has a useful hint.

 

[anuttarasammyak]

I don't think that is necessary. To make sure that the comparison between the two cases is meaningful, you have to assume that the same mass of water at 100 oC is converted to steam at 100 oC.

Volume of 1 kg steam of 100 oC with no dispersion in 100 oC atmosphere seems depend on pressure of atmosphere as variable forgetting that 100 oC is water boiling temperature at standard atmosphere pressure.