Question

A vessel with an unknown volume is filled with 10 kg of water at 90oC. Inspection of the vessel at equilibrium shows that 8 kg of the wateris in the liquid state. What is the pressure in the vessel, and what is the volume of the vessel

Answer 1

Hey there!

In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).

Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:

$x=\frac{m_{steam}}{m_{total}}$

Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:

$x=\frac{2kg}{10kg} =0.20$

Now, at this temperature and pressure, the volume of a saturated vapor is  2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:

$v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg$

This means that the volume of the container will be:

$V=10kg*0.4727m^3/kg\\\\V=4.73m^3$

Regards!