What is the State of Water in a Rigid Container at 120oC?

[Monsterboy]

Homework Statement

A rigid container of volume 0.5 m³ contains 1.0 kg of water at 120^oC (v_f= 0.00106 m³/kg ,v_g=0.8908 m³/kg )
v-specific volume (v_f is v of saturated liquid and v_g is v of saturated vapour)
The state of water is
(a) compressed liquid
(b) saturated liquid
(c) a mixture of saturated liquid and saturated vapor.
(d) super-heated vapor

Homework Equations

v = v_f + x(v_g- v_f)
x- dryness fraction

The Attempt at a Solution

In order to know whether the water can exist in vapor state in the container at 120^oC , i need to find the pressure? because pressure decides the boiling point.

If i consider dryness fraction of zero then the volume occupied will be 0.00106 m³ right ? given that the volume of the container is 0.5 m³ certainly the water is not compressed so option (a) is ruled out.

In order to rule out option (b) and (d) , i need to know the pressure right ? i don't know how to proceed.

The answer given to me is (c) but i want to know how to rule out (b) and (d).

 

[Bystander]

0.00106 m³ is how much mass?

 

[ehild]

Homework Statement

A rigid container of volume 0.5 m³ contains 1.0 kg of water at 120^oC (v_f= 0.00106 m³/kg ,v_g=0.8908 m³/kg )
v-specific volume (v_f is v of saturated liquid and v_g is v of saturated vapour)
The state of water is
(a) compressed liquid
(b) saturated liquid
(c) a mixture of saturated liquid and saturated vapor.
(d) super-heated vapor

Homework Equations

v = v_f + x(v_g- v_f)
x- dryness fraction

The Attempt at a Solution

In order to know whether the water can exist in vapor state in the container at 120^oC , i need to find the pressure? because pressure decides the boiling point.

If i consider dryness fraction of zero then the volume occupied will be 0.00106 m³ right ? given that the volume of the container is 0.5 m³ certainly the water is not compressed so option (a) is ruled out.

In order to rule out option (b) and (d) , i need to know the pressure right ? i don't know how to proceed.

The answer given to me is (c) but i want to know how to rule out (b) and (d).

The volume of liquid water is negligible with respect to the volume of the container. You are given the specific volume of the vapour at 120 °C. How much is the mass of the vapour in the container? Is it less or greater than the total mass of water?

 

[Monsterboy]

0.00106 m³ is how much mass?

According to the values given, the total amount of water in the container is 1kg ,if i consider dryness fraction as zero , i get specific volume as 0.00106 m³/kg ,so one kg of saturated liquid water has a volume of 0.00106 m³ right ?

At 120^oC i don't think it is possible for only liquid water to exist in saturated form in such a large volume of the system because the rest of it will be vacuum then right ? can i rule out option(b) this way ?

 

[Monsterboy]

How much is the mass of the vapour in the container? Is it less or greater than the total mass of water?

The dryness fraction is not given and i don't know how to find it. The total mass of the water (either liquid or vapour or both) in the system is given as 1kg.

 

[Bystander]

Is 0.00106 m³ more or less than 1 kg?

 

[Monsterboy]

Is 0.00106 m³ more or less than 1 kg?

I am not able to get what you are asking , i said 1kg of saturated liquid water occupies a volume of 0.00106 m³ (assuming dryness fraction as zero)

 

[ehild]

"A rigid container of volume 0.5 m³ contains 1.0 kg of water at 120^oC (v_f= 0.00106 m³/kg ,v_g=0.8908 m³/kg )
v-specific volume (v_f is v of saturated liquid and v_g is v of saturated vapour)"

Forget the dryness factor.
v_f = 0.00106 m³/kg is the specific volume of the water. So 1 kg water would occupy 0.00106 m³, much less than the volume of the container.
So the volume of 0.5 m³ is mostly filled in by the saturated vapour. Knowing the specific volume of the vapour, v_g=0.8908 m³/kg, what is the mass of 0.5 m³ of it?
Then how much water is in liquid state?

 

[Chestermiller]

Why don't you just solve for x?

Chet

 

[ehild]

Why don't you just solve for x?

Chet

The OP said x was the dryness factor. I read that the dryness factor is something characterizing the steam, the amount of water droplets in it.

 

[Chestermiller]

The OP said x was the dryness factor. I read that the dryness factor is something characterizing the steam, the amount of water droplets in it.

x is the mass fraction of the water that is in the vapor phase, and v is the mass-average specific volume of the liquid and vapor.

 

[ehild]

Is

x is the mass fraction of the water that is in the vapor phase, and v is the mass-average specific volume of the liquid and vapor.

There is no liquid water phase on the bottom of the container? The whole water is in the form of wet steam?

 

[Chestermiller]

Is

There is no liquid water phase on the bottom of the container? The whole water is in the form of wet steam?

If there's gravity, then there is liquid water in the bottom of the container and water vapor in the head space (at equilibrium). If there is no gravity (or if all the liquid water hasn't coalesced yet), then there can be liquid water drops in the vapor phase. But the vapor is gas, and the liquid is liquid. For that equation to apply, it doesn't matter where the liquid water and the water vapor are situated in the container.

Chet

 

[Monsterboy]

Forget the dryness factor.
v_f = 0.00106 m³/kg is the specific volume of the water. So 1 kg water would occupy 0.00106 m³, much less than the volume of the container.
So the volume of 0.5 m³ is mostly filled in by the saturated vapour. Knowing the specific volume of the vapour, v_g=0.8908 m³/kg, what is the mass of 0.5 m³ of it?
Then how much water is in liquid state?

I don't think the rest of the volume is filled by saturated vapour ,because the total mass in the system itself is 1 kg only , if 1 kg of saturated liquid water is sitting at the bottom occupying a volume of 0.00106 m³ ,there is nothing else left to occupy the remaining volume in the container!

 

[Chestermiller]

I don't think the rest of the volume is filled by saturated vapour ,because the total mass in the system itself is 1 kg only , if 1 kg of saturated liquid water is sitting at the bottom occupying a volume of 0.00106 m³ ,there is nothing else left to occupy the remaining volume in the container!

This is not correct. Just solve for x with your equation and tell us what you get. This will tell you the precise mass of water in the vapor phase and the precise mass of water in the liquid phase.

 

[Monsterboy]

This is not correct.
Just solve for x with your equation and tell us what you get. This will tell you the precise mass of water in the vapor phase and the precise mass of water in the liquid phase.

I got x=0.5607 ,so the answer is (c) ,is that all ?

 

[Chestermiller]

I got x=0.5607 ,so the answer is (c) ,is that all ?

You don't get off that easily now. Now that you have this result, tell us the mass of liquid, the mass of vapor, the volume of vapor, and the volume of liquid in the 0.5 m³ container.

 

[Monsterboy]

You don't get off that easily now. Now that you have this result, tell us the mass of liquid, the mass of vapor, the volume of vapor, and the volume of liquid in the 0.5 m³ container.

Mass of vapour =0.56 kg
mass of liquid =0.44 kg
Volume of vapour = 0.0005936 m³
volume of liquid = 0.391952 m³

the volume numbers don't sound right but that's what i got.

 

[Chestermiller]

Mass of vapour =0.56 kg
mass of liquid =0.44 kg
Volume of vapour = 0.0005936 m³
volume of liquid = 0.391952 m³

the volume numbers don't sound right but that's what i got.

The volume numbers are not right. Show us how you got these numbers. They have to add up to 0.5 m³.

 

[Monsterboy]

The volume numbers are not right. Show us how you got these numbers. They have to add up to 0.5 m³.

I just multiplied the mass of the liquid and the specific volume v_f to get the volume of the liquid ,i did similarly for the volume of the vapour with v_g.

 

[Chestermiller]

I just multiplied the mass of the liquid and the specific volume v_f to get the volume of the liquid ,i did similarly for the volume of the vapour.

It looks like you multiplied the mass of the liquid by the specific volume of the vapor and the mass of vapor by the specific volume of the liquid.

Chet

 

[Monsterboy]

It looks like you multiplied the mass of the liquid by the specific volume of the vapor and the mass of vapor by the specific volume of the liquid.

Chet

Oops !
Volume of vapour = 0.4988 m³
volume of liquid =0.0004664 m³

Thanks for the help !