# Will Heating a Sealed Container Cause All Liquid to Convert to Vapor?

• I
• deepanshu_ch
In summary, the conversation discussed the process of heating a closed container with liquid present and the conversion of the liquid into vapor. The participants also touched on the concept of dynamic equilibrium between the two phases and how the rates of evaporation and condensation are equal in this state. The conversation concluded with the suggestion to use the equation derived to perform calculations for a specific scenario.
deepanshu_ch
Please can one explain what is exactly gonna happen if we have closed container and liquid is present in it and we heat from bottom ? Then whole liquid will converted into vapour or not ? But during the phase change both system in dynamic equilibrium. If both phases are in dynamic equilibrium then how the liquid completely converted into vapour

We can quantify exactly what is going to happen by making use of the condition that the sum of liquid volume plus vapor volume is equal to the constant volume of the container:
$$V_L+V_V=V$$where $$V_L=m_Lv_L$$and$$V_V=m_Vv_V$$with ##m_L## representing the mass of liquid in the container, ##m_V## representing mass of vapor in the container, ##v_L## representing the specific volume of liquid water at the contents temperature, and ##v_V## representing the specific volume of water vapor at the contents temperature. If m represents the total amount of water in the container and x is taken to represent the mass fraction water vapor in the container, then $$m_L=m(1-x)$$and$$m_V=mx$$If we combine all these equations,, we obtain:$$m(1-x)v_L+mxv_V=V$$or, rearranging,$$v_L+x(v_V-v_L)=\frac{V}{m}$$or, rearranging further,$$x=\frac{\bar{v}-v_L}{v_V-v_L}$$where $$\bar{v}=\frac{V}{m}$$OK so far?

Is vL can be zero on continuously heating ?

deepanshu_ch said:
Is vL can be zero on continuously heating ?
vL, the specific volume of saturated liquid water, is a physical property of water that is a function only of the temperature T; you can look up this variation with temperature in the "steam tables." If has a positive value at all temperatures up to the critical temperature (at which it is equal to the specific volume of the saturated vapor).

deepanshu_ch

Chestermiller said:
vL, the specific volume of saturated liquid water, is a physical property of water that is a function only of the temperature T; you can look up this variation with temperature in the "steam tables." If has a positive value at all temperatures up to the critical temperature (at which it is equal to the specific volume of the saturated vapor).
Look at above image where completely water is converted into water vapour. Is this case of closed container or open container ?

deepanshu_ch said:
Look at above image where completely water is converted into water vapour. Is this case of closed container or open container ?
What do you think would happen if you had a single drop of liquid water present in a huge closed container and you heated to raise the temperature? Do you think the drop would not evaporate? Are you ready to do some actual calculations using the equation I derived for a closed container?

deepanshu_ch
Chestermiller said:
What do you think would happen if you had a single drop of liquid water present in a huge closed container and you heated to raise the temperature? Do you think the drop would not evaporate? Are you ready to do some actual calculations using the equation I derived for a closed container?
I get it
But how equilibrium set up between liquid and vapour
I m trying to explain if I m wrong Correct me
When we heat vapour starting form because molecules at liquid surface have enough energy to leave the liquid then quantity will start increasing in vapour phase . After some time the dynamic equilibrium set up between the liquid and and vapour phase . As we are continuously heating at any instant so molecules leaving the liquid phase will increase . Then rate of condensation will also increase.
At conclusion can we said
Dynamic equilibrium set up between phases
Both rate of condensation and evaporation will increase and equal and this cycle repeat

Last edited:
deepanshu_ch said:
I get it
But how equilibrium set up between liquid and vapour
I m trying to explain if I m wrong Correct me
When we heat vapour starting form because molecules at liquid surface have enough energy to leave the liquid then quantity will start increasing in vapour phase . After some time the dynamic equilibrium set up between the liquid and and vapour phase . As we are continuously heating at any instant so molecules leaving the liquid phase will increase . Then rate of condensation will also increase.
At conclusion can we said
Dynamic equilibrium set up between phases
Both rate of condensation and evaporation will increase and equal and this cycle repeat
That sounds about right to me. I the heating is relatively slow, the system is very close to equilibrium at any time. Even if it is not, once you stop heating, the system will re-equilibrate at a new temperature.

Below is an abbreviated steam table of specific volume (liters/kg) of saturated liquid water and saturated water vapor as a function of temperature (C). I would like you to do some calculations using the equation I developed in Post #2.

Assume we have a rigid container of volume 10 liters holding 1 kg of water (liquid + vapor). What is the value of ##\bar{v}=V/m## for this situation (liters/kg)? What is the mass fraction water vapor x in the container at each of the temperatures in the table? Make a graph of x vs T.

deepanshu_ch
deepanshu_ch said:
evaporation

Chestermiller said:
Below is an abbreviated steam table of specific volume (liters/kg) of saturated liquid water and saturated water vapor as a function of temperature (C). I would like you to do some calculations using the equation I developed in Post #2.

View attachment 328044

Assume we have a rigid container of volume 10 liters holding 1 kg of water (liquid + vapor). What is the value of ##\bar{v}=V/m## for this situation (liters/kg)? What is the mass fraction water vapor x in the container at each of the temperatures in the table? Make a graph of x vs T.

Chestermiller said:
Below is an abbreviated steam table of specific volume (liters/kg) of saturated liquid water and saturated water vapor as a function of temperature (C). I would like you to do some calculations using the equation I developed in Post #2.

View attachment 328044

Assume we have a rigid container of volume 10 liters holding 1 kg of water (liquid + vapor). What is the value of ##\bar{v}=V/m## for this situation (liters/kg)? What is the mass fraction water vapor x in the container at each of the temperatures in the table? Make a graph of x vs T.
How can be fraction greater than 1 ?

It means that all the liquid water has evaporated, you no longer have vapor-liquid equilibrium, and the container is filled with superheated vapor (below the equilibrium vapor pressure).

deepanshu_ch
Ohh... I get it . Thanks for explaining .

deepanshu_ch said:
Ohh... I get it . Thanks for explaining .
Not so fast. Now consider 1 kg of water in a 1.25 liter container.

Chestermiller said:
Not so fast. Now consider 1 kg of water in a 1.25 liter container.

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container gonna be blast at negative value of x .
Am I right ?

deepanshu_ch said:
container gonna be blast at negative value of x .
Am I right ?
No. We are assuming that the container is infinitely rigid.

It means that there is only liquid water remaining in the container under very high pressure; the pressure is above the equilibrium vapor pressure of water at the temperature, and the specific volume is less than the equilibrium specific volume of liquid water at the temperature.

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deepanshu_ch

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