Using the 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.

 

[kuruman]

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.

For a meaningful comparison, one needs a controlled process. What I have in mind is the thought experiment shown on the right.

Start with two identical cylindrical vessels containing identical amounts of water at 100^oC.

Cap both vessels with a piston of negligible mass.

Put some extra weight on top of the piston on the right. This increases the pressure at the surface of the water and the boiling point of the water in the cylinder on the right to, say, 102^oC.

Add controlled amounts of heat Q and Q' as shown to generate steam that displaces both pistons by the same infinitesimal amount $\Delta V.$

For each case, write the first law for the steam expansion by $\Delta V.$

Draw your conclusions in the limit $\Delta V \rightarrow 0.$

On edit
The initial figure and text has been altered to incorporate the comments by @Lnewqban in post #14.

 

[anuttarasammyak]

Good design. Say steam is ideal gas
V=nRT/p
Say n and T are fixed, V depends on p which depends on weight.

 

[Lnewqban]

View attachment 374231
Put some extra weight on top of the piston on the right. This increases the pressure at the surface of the water.

I believe that there should not be steam formation below the piston on the right when water temperature is only 100 °C.
Increasing the pressure at the surface of the water, also increases the boiling temperature.

 

[kuruman]

I believe that there should not be steam formation below the piston on the right when water temperature is only 100 °C.
Increasing the pressure at the surface of the water, also increases the boiling temperature.

You are absolutely correct. I was a bit hasty cutting and pasting labels in the figure. The original post has been edited.

 

[anuttarasammyak]

The original post has been edited.

Q‘=Q+W > Q
where W is work done to the water liquid-steam system in adiabatic compression process from Figure 3 to 4 where number is from the left. I hope this is an answer to OP.

[EDIT]

draw your conclusions in the limit

It may make the discussion simpler if we assume that 1 kg of water is completely converted into steam of the boiling points by heat Q or Q′.

I think that temperature of Figure 2 should be 102 so that Q' do not include heating water to boiling temperature. If so initial internal energy is different and the above inequality becomes ambiguous.　　　I asked AI to show some data.

Latent Heat of vaporization decreases as pressure increases. Now I am not sure whether 1st law of thermodynamics is enough to explain it or not.

 

[songoku]

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

Increasing the pressure for same ΔV will increase the work done by the system so the work done on the system will decrease (will be more negative)

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.

ΔU = change in internal energy of water
ΔKE = change in kinetic energy of water molecules
ΔPE = change in potential energy of water molecules

 

[kuruman]

Increasing the pressure for same ΔV will increase the work done by the system so the work done on the system will decrease (will be more negative)


ΔU = change in internal energy of water
ΔKE = change in kinetic energy of water molecules
ΔPE = change in potential energy of water molecules

These are just names. What specific expressions for each one of them do you think you can use to relate them to relevant quantities in the problem?

 

[songoku]

These are just names. What specific expressions for each one of them do you think you can use to relate them to relevant quantities in the problem?

I am not sure what you mean.

ΔU = Q + W where W is work done on the system

Q = m L where L is the specific latent heat of vaporization and is the one we want to find out

ΔU = ΔKE + ΔPE = 1/2 m (v₂² - v₁²) + ΔPE
Sorry I don't know the formula of ΔPE for this case. I just know PE in this case is related to separation between molecules.

Is that what you mean?

 

[anuttarasammyak]

Latent Heat of vaporization decreases as pressure increases. Now I am not sure whether 1st law of thermodynamics is enough to explain it or not.

As far as I see in the web and AI answers they use Clapeyron Clasusius equation to explain it. I wonder if it is what expected by the homework teacher.

 

[kuruman]

As far as I see in the web and AI answers they use Clapeyron Clasusius equation to explain it. I wonder if it is what expected by the homework teacher.

I don't think that one has to go to Clausius-Clapeyron for this. It looks like energy conservation to me as the OP started doing. You add heat ΔQ to water at the boiling point and you end up with most of the water at the boiling point, with some having been converted to steam under pressure. The question is where did ΔQ go?

 

[anuttarasammyak]

You add heat ΔQ to water at the boiling point and you end up with most of the water at the boiling point, with some having been converted to steam under pressure. The question is where did ΔQ go?

I will try. For 1kg liquid water at boiling temperature T under constant pressure p to become steam gas completely, we provide heat which is divided into latent heat and work the system does.

$$Q=L+W=L+p(V_g-V_l)$$

In approximation that $V_l << V_g$ for occupied volumes of liquid water and steam gas, and steam is an ideal gas

$$Q \approx L+p\frac{nRT}{p}=L+nRT$$

where $n=\frac{1000}{18}=55.6\ mol$

If I am right up to here how should I do further ?

[EDIT] Another approach
For 1 kg or 1 mol water. Say coexistence line is f(p,T)=0
$$f(p_1,T_1)=0, f(p_2,T_2)=0$$
$$\triangle p := p_2-p_1 >0 $$
$$\triangle T := T_2-T_1 > 0*$$
Both of them are taken small. An enthalpy difference of the two states in first order
$$H(p_2,T_2;gas)-H(p_1,T_1;liquid)$$
where H is enthalpy and L is latent heat. In first order it is
$$=L(p_1,T_1)+Cp(p_1;gas)\triangle T+ V_g\triangle p $$
by the route vaporizing first, heating second and pressurising third, and
$$=V_l \triangle p+Cp(p_2;liquid)\triangle T+L(p_2,T_2) $$
by the route pressurising first, heating second and vaporizing third, where V_g, V_l are voulumes of gas and liquid in (p,T) around. By equating the result of the two routes
$$L(p_2,T_2)-L(p_1,T_1)$$
$$=-[Cp(p_2;liquid)-Cp(p_1;gas)]\triangle T+(V_g-V_l)\triangle p+second\ order$$
$$=-[Cp(liquid)-Cp(gas)]\triangle T+(V_g-V_l)\triangle p+second\ order$$
all under same pressure p=p_1 or p_2, T=T_1 or T_2 on the coexistence line. It is Kirchfoff's equaiton itself or its variation.

$V_g-V_l>0$. In our experience (not from theory) for water
$$Cp(p;liquid)-Cp(p;gas)>0$$
and
$$\frac{\triangle p}{\triangle T}|_{along\ the\ coexistence\ line}>0$$
as already assumed*. Thus signature of $L(p_2,T_2)-L(p_1,T_1)$ is determined by
which of the first minus term and the second plus term dominates.

Clausius-Clapeyron equation suggests that the first minus term dominates thus latent heat decdreases as pressure or temperature increases along the coexistence line.

 

[kuruman]

First of all, @songoku who posted the problem; we are here to help along a path to the solution.

Secondly, you can see for yourself that your last equation is problematic in that the heat that you calculated is independent of pressure. In other words, whether the pressure is atmospheric or greater than atmospheric, the result will be the same!

The heat that you calculated is what is required to boil away a certain amount of water with the resulting steam doing no work on anything.

 

[songoku]

As far as I see in the web and AI answers they use Clapeyron Clasusius equation to explain it. I wonder if it is what expected by the homework teacher.

I do not learn Clapeyron Clausius equation

@kuruman what is the mistake in post #19?

 

[kuruman]

I do not learn Clapeyron Clausius equation

@kuruman what is the mistake in post #19?

You don't need Clausius-Clapeyron equation for this.

In post #19 it's more a misuse than a mistake. In this problem you have an amount of water just at the boiling point at atmospheric pressure. You add a little bit of heat Q and a certain number of water molecules come out as steam which you can assume is at the boiling temperature of the water. You do the same with water just at the boiling temperature but at pressure higher than atmospheric. That's the picture suggested by the statement of the problem.

In post #19 you wrote the equation

ΔU = ΔKE + ΔPE = 1/2 m (v₂² - v₁²) + ΔPE

What are the meanings of the symbols in this equation in terms of the picture suggested by the problem? How relevant are they? What is ΔPE in terms of the picture? How does the pressure enter in the picture?

 

[songoku]

In post #19 you wrote the equation

ΔU = ΔKE + ΔPE = 1/2 m (v₂² - v₁²) + ΔPE

What are the meanings of the symbols in this equation in terms of the picture suggested by the problem? How relevant are they? What is ΔPE in terms of the picture? How does the pressure enter in the picture?

Case (1) = pressure is equal to atmospheric pressure
Case (2) = pressure is higher than atmospheric pressure

Pressure higher than atmospheric pressure will increase the boiling point of water. Water molecules in case (2) would have higher KE than in case (1) because they are at higher temperature but ΔKE for both cases will be zero since there is no temperature change during boiling (assuming the initial temperature of water is at boiling point of each case)

PE is related to separation between molecules. Since we assume ΔV is the same for both cases, it means ΔPE will be the same for both cases so ΔU₁ = ΔU₂

Am I correct so far?

 

[kuruman]

Since we assume ΔV is the same for both cases, it means ΔPE will be the same for both cases so ΔU1 = ΔU2

Why does it mean that? What exactly are those changes in internal energy? The change from what point of the process to what other point of the process?

 

[songoku]

Why does it mean that? What exactly are those changes in internal energy? The change from what point of the process to what other point of the process?

The change from the condition where the object is water at 100^oC to the condition where the object is vaopur at 100^oC

Since internal energy is the sum of kinetic and potential energy and for both cases the change in KE and PE are the same so the change in internal energy is also the same