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Determine the condition that describes the state of water at 191°C and 200 cm3/g.

Determine the condition that describes the state of water at 191°C and 200 cm3/g.

    1. Saturated vapor

    2. Saturated liquid

    3. 2-phase

    4. Superheated gas

    5. Compressed liquid

Solutions

Expert Solution

We know that the volume of water is 1 L/kg
or 1 ml per gram.
Since the given value is 200 cm^3/ml or 200 ml per gram of water.

It is obvious that we have water is not in the form of liquid only however we can take one check again.

We can use humidity chart for the purpose.
Specific volume .
Specif volume of water is 1.13 cm^3 per gram at this temperature and that of vapor can be calculated by taking ideal gas assumption.

Density =PM/RT
Here P is the pressure.
M is the molecular weight.
R is the universal gas constant.
T is the temperature in K.
Density
= 101325*18*0.001/(8.314*(273+191))
=101325*18*0.001/8.314*464
=0.47 kg per m^3
470 gram occupies 1 m^3.
1 gram occupies 1000000 cm^3/470
=2127 cm^3 / gram

We can see that given value of the specific volume of water is between the value obtained for gas and liquid. So, at this point we have a mixture of liquid and vapor.

So, option C , 2 -phase is correct answer.

Amanda K answered 2 years ago
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