Question

The specific volume of 5 kg of water vapor at 1.5 MPa, 440°C is 0.2160 m3 /kg. Determine (a) the volume, in m3 , occupied by the water vapor, (b) the amount of water vapor present, in gram moles, and (c) the number of molecules.

Answer 1

Answer:

(a) Volume=1.08 m³

(b) η=273.3 moles

(c) n=1.646×10²⁶ molecules

Explanation:

Given data

Mass m=5 kg

Pressure P=1.5 MPa

Temperature t=440°C=(440+273)K=713 K

Specific volume v=0.2160 m³/kg

To find

(a) Volume Occupied

(b) Amount of water vapor grams per mole

(c) Number of molecules

Solution

Let the system is in equilibrium and vapor is ideal gas

For part (a) volume occupied

$volume=m_{mass} *v_{specific-volume}\\ volume=(5kg)*(0.216m^{3}/kg )\\volume=1.08m^{3}$

For part (b) amount of water vapor grams per mole

Apply ideal gas law

$PV=n.R.T\\n=\frac{PV}{R.T}\\ n=\frac{(1.5*10^{6} )*(1.08)}{(8.314)(713)}\\ n=273.3 moles$

For (c) number of molecules

$n_{molecules}=n*N_{Avogadro-number}\\n_{molecules}=273.3*(6.02214*10^{23} )\\ n_{molecules}=1.646*10^{26}molecules$