Thermodynamics pure substances problem

In summary: This equation is saying that the mass of water vapor is equal to the mass of total water multiplied by the mass fraction of water vapor. So, in this case, the 1 comes from the fact that there is only one unit of water vapor in the whole tank.
  • #1
hazzz
7
0

Homework Statement


Hellooo, so this is the question that i have:

A 0.5-m3 rigid tank initially contained a saturated liquid-vapor mixture of water at 140 °C is now heated until the mixture reaches the critical state. Determine the mass and the volume of liquid before the heating process

Homework Equations


v = vf + x*vfg
x = mvapor/mtotal
v = V/m

The Attempt at a Solution


I do not understand how to find the mass before the heating process :(
I found the X which is 0,004086
vf at 140C is 0,001080m3/kg
vg at 140C is 0,5089m3/kg

the final answer should be mass = 157,83 kg
 
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  • #2
hazzz said:

Homework Statement


Hellooo, so this is the question that i have:

A 0.5-m3 rigid tank initially contained a saturated liquid-vapor mixture of water at 140 °C is now heated until the mixture reaches the critical state. Determine the mass and the volume of liquid before the heating process

Homework Equations


v = vf + x*vfg
x = mvapor/mtotal
v = V/m

The Attempt at a Solution


I do not understand how to find the mass before the heating process :(
I found the X which is 0,004086
vf at 140C is 0,001080m3/kg
vg at 140C is 0,5089m3/kg

the final answer should be mass = 157,83 kg
There seems to be information missing. Have you provided the complete problem statement?
 
  • #3
Chestermiller said:
There seems to be information missing. Have you provided the complete problem statement?
yes everything is there, unless if you do not have access to the saturated water tables! :O For the critical state, the value of v is 0,003155m3/kg
 
  • #4
hazzz said:
yes everything is there, unless if you do not have access to the saturated water tables! :O For the critical state, the value of v is 0,003155m3/kg
If it can be heated to exactly the critical state, then you can use this specific volume value for the critical state to calculate the total mass of water in the tank. What is that total mass of water in the tank?

EDIT: Wait a second. If you determined X, then you must already have done this. X is the mass fraction of water vapor. What is the mass fraction of liquid water? Given that you know the total mass of water in the tank and the mass fraction of liquid water, what is the mass of liquid water?

Chet
 
  • #5
Chestermiller said:
If it can be heated to exactly the critical state, then you can use this specific volume value for the critical state to calculate the total mass of water in the tank. What is that total mass of water in the tank?

Chet
Yes i got 158,48kg :) but what i am suppose to do after finding this value? Is this the total mass of water in the tank after or before the heating process ?
 
  • #6
Chestermiller said:
If it can be heated to exactly the critical state, then you can use this specific volume value for the critical state to calculate the total mass of water in the tank. What is that total mass of water in the tank?

EDIT: Wait a second. If you determined X, then you must already have done this. X is the mass fraction of water vapor. What is the mass fraction of liquid water? Given that you know the total mass of water in the tank and the mass fraction of liquid water, what is the mass of liquid water?

Chet
I don't know :( In the solution book they wrote:

mf = (1-xf) * mtotal
where total = 158,48 kg but i don't know where they took 1-xf and what is xf
 
  • #7
hazzz said:
Yes i got 158,48kg :) but what i am suppose to do after finding this value? Is this the total mass of water in the tank after or before the heating process ?
It is both. The mass of water in the tank has not changed.
 
  • #8
hazzz said:
I don't know :( In the solution book they wrote:

mf = (1-xf) * mtotal
where total = 158,48 kg but i don't know where they took 1-xf and what is xf
xf is what you called X in your original post. It is the mass fraction of water vapor in the tank. So, 1-xf is the mass fraction of liquid water in the tank. So, if you know the total mass of water in the tank and the mass fraction of this water that is liquid, you can find the mass of liquid water.
 
  • #9
Chestermiller said:
xf is what you called X in your original post. It is the mass fraction of water vapor in the tank. So, 1-xf is the mass fraction of liquid water in the tank. So, if you know the total mass of water in the tank and the mass fraction of this water that is liquid, you can find the mass of liquid water.
but where does the 1 come from? I thought the equation was X = mvapor/mtotal so if my X is equal to 0,004086 and my mtotal is 158,48kg how do they get 157,83kg ?
 
  • #10
hazzz said:
but where does the 1 come from? I thought the equation was X = mvapor/mtotal so if my X is equal to 0,004086 and my mtotal is 158,48kg how do they get 157,83kg ?
The sum of the mass fractions has to add up to 1. $$\frac{mvapor}{mtotal}+\frac{mliquid}{mtotal}= \frac{(mvapor + mliquid)}{mtotal}=1$$If the mass fraction of vapor is 0.004086, then the mass fraction of liquid must be 1-0.004086.
 
  • #11
Chestermiller said:
The sum of the mass fractions has to add up to 1. $$\frac{mvapor}{mtotal}+\frac{mliquid}{mtotal}= \frac{(mvapor + mliquid)}{mtotal}=1$$If the mass fraction of vapor is 0.004086, then the mass fraction of liquid must be 1-0.004086.
OHHHH okay! I understand now! :D Thank you very much! You are the best!
 

1. What is the definition of a "pure substance" in thermodynamics?

A pure substance in thermodynamics is a material that has a constant chemical composition throughout and exhibits the same thermodynamic properties regardless of its physical state (solid, liquid, or gas).

2. How do you determine the state of a pure substance in a thermodynamics problem?

The state of a pure substance is determined by two intensive properties: temperature and specific volume. This can be represented on a thermodynamic diagram, such as a T-v or P-v diagram.

3. What is the difference between an ideal gas and a real gas in thermodynamics?

An ideal gas is a hypothetical gas that follows the ideal gas law, while a real gas deviates from this law due to intermolecular forces. Real gases also exhibit non-ideal behavior at high pressures and low temperatures.

4. How do you calculate the work done by a pure substance in a thermodynamics problem?

The work done by a pure substance can be calculated by integrating the pressure-volume (P-v) curve on a thermodynamic diagram. This is represented by the area under the curve.

5. What is the significance of the first and second laws of thermodynamics in pure substance problems?

The first law of thermodynamics states that energy cannot be created or destroyed, only converted from one form to another. The second law states that the total entropy of a closed system will always increase over time. These laws are important in determining the energy and entropy changes in a pure substance system.

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