# Thermodynamic Challenge Problem

• Chestermiller
In summary, the conversation discusses a thermodynamic problem involving a 1 liter cylinder with liquid water and air at initial conditions of 298 K and 1 bar. The external pressure is increased to 10 bars, causing adiabatic compression and resulting in some of the water evaporating. The final equilibrium state is to be determined for both a sudden and gradual increase in pressure, with varying amounts of initial liquid water. The conversation includes hints for solving the problem using equations for internal energy and entropy, and also mentions the possibility of using steam tables as a resource.
Chestermiller
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I've seen a thread posted on another forum which described a thermodynamic situation that captured my interest, so I though I would introduce a challenge problem on it. The other forum was not able to adequately specify or address how to approach a problem like this. I know how to solve this problem, but let's see how others at Physics Forums might approach it.

PROBLEM STATEMENT: I have a 1 liter cylinder with a massless, frictionless piston. There is some liquid water in the cylinder, with air above it. The temperature is initially 298 K, and the system pressure matches the outside pressure of 1 bar. I increase the external pressure to 10 bars (a) suddenly or (b) gradually/reversibly. The system is adiabatic. The increase in temperature from adiabatic compression of the gas phase causes some of the water to evaporate. In the final equilibrium state of the system, determine the final temperature, volume, and mass of liquid water for scenarios (a) and (b). First consider the case where there is exactly enough liquid water initially, such that, at final equilibrium, all of the water has just barely evaporated. Then consider cases where there is initially twice this amount of liquid water and half this amount for each of the two scenarios.

Allowable assumptions are that
1. The gas phase can be treated as an ideal gas mixture
2. Negligible air dissolves in the water

anorlunda and berkeman
It looks like no one is interested in responding to this yet. So, here are some hints:

1. Call ##n_w## the initial number of moles of liquid water in the cylinder
2. Call ##\Delta n_w## the increase in the number of moles of water in the gas phase at final equilibrium
3. Call ##\lambda## the change in internal energy per mole upon vaporization of water at the initial temperature
4. Let ##v_w## represent the specific molar volume of liquid water at the initial conditions
5. How many moles of air are in the cylinder?

I'll try to do case (b) first.
Since it is done reversibly, I know $\Delta S = 0$.
$$P_1 = 1 \rm{bar}$$ $$T_1 = 298 K$$, $$P_2 = 10 \rm{bar}$$.
Since some work $W$ is being done on the system, $$U_1 + W = U_2$$.
Where, $$U_1 = n_aU_{air}(T_1, P_1) + n_wU_w (T_1,P_1)$$.
$$U_2 = n_aU_{air}(T_2,y_aP_2) + (\Delta n_w)U_{w,g} (T_2, y_wP_2) + (n_w-\Delta n_w) U_{w,l}(T_2,P_2)$$
Where $y_w$ is the mole fraction of steam in the vapor phase and $y_a$ is the mole fraction of air in the vapor phase.

We can write a similar expression for entropy for state 1 and state 2.
$$\Delta S = 0$$
$$S_1 = n_aS_{air}(T_1, P_1) + n_wS_w (T_1,P_1)$$
$$S_2 = n_aS_{air}(T_2,y_aP_2) + (\Delta n_w)S_{w,g} (T_2, y_wP_2) + (n_w-\Delta n_w) S_{w,l}(T_2,P_2)$$

I don't see how using the heat of vaporization applies to this situation...

Last edited:
berkeman
Sat D said:
I'll try to do case (b) first.
Since it is done reversibly, I know $\Delta S = 0$.
$$P_1 = 1 \rm{bar}$$ $$T_1 = 298 K$$, $$P_2 = 10 \rm{bar}$$.
Since some work $W$ is being done on the system, $$U_1 + W = U_2$$.
Where, $$U_1 = n_aU_{air}(T_1, P_1) + n_wU_w (T_1,P_1)$$.
$$U_2 = n_aU_{air}(T_2,y_aP_2) + (\Delta n_w)U_{w,g} (T_2, y_wP_2) + (n_w-\Delta n_w) U_{w,l}(T_2,P_2)$$
Where $y_w$ is the mole fraction of steam in the vapor phase and $y_a$ is the mole fraction of air in the vapor phase.

We can write a similar expression for entropy for state 1 and state 2.
$$\Delta S = 0$$
$$S_1 = n_aS_{air}(T_1, P_1) + n_wS_w (T_1,P_1)$$
$$S_2 = n_aS_{air}(T_2,y_aP_2) + (\Delta n_w)S_{w,g} (T_2, y_wP_2) + (n_w-\Delta n_w) S_{w,l}(T_2,P_2)$$

I don't see how using the heat of vaporization applies to this situation...
This looks pretty good so far. The heat of vaporization comes into play in determining ##S_{w,g}(T_2, y_wP_2)##. Can you figure out how to express this in terms of ##S_{w,l}(T_1, P_1)##, the heat of vaporization, and the heat capacity of water vapor?

Would we simply be using the ideal gas approximation for entropy changes here?
So $$S_{w,g}(T_2,y_wP_2) = \frac{\lambda}{T_1} + C_p\ln\frac{T_2}{T_1} + R\ln \frac{P_2}{P_1}+S_{w,l}(T_1,P_1)$$
Is that right?

My question is, if I looked up the value of the entropy of steam at ##T_2, y_wP_2## in a steam table, they should be reasonably close?

anorlunda
Sat D said:
Would we simply be using the ideal gas approximation for entropy changes here?
So $$S_{w,g}(T_2,y_wP_2) = \frac{\lambda}{T_1} + C_p\ln\frac{T_2}{T_1} + R\ln \frac{P_2}{P_1}+S_{w,l}(T_1,P_1)$$
Is that right?

My question is, if I looked up the value of the entropy of steam at ##T_2, y_wP_2## in a steam table, they should be reasonably close?
Yes, except that there should be a minus sign in front of the pressure term. And I didn't say anything in the problem statement about not being able to use the steam tables, so that's OK too. And yes, the values should be reasonably close.

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