Given:
mass of water, m = 2000 kg
temperature, T = [tex]30^{\circ}C[/tex] = 303 K
extacted mass of water = 100 kg
Atmospheric pressure, P = 101.325 kPa
Solution:
a) Using Ideal gas equation:
PV = m[tex]\bar{R}[/tex]T (1)
where,
V = volume
m = mass of water
P = atmospheric pressure
[tex]\bar{R} = \frac{R}{M} [/tex]
R= Rydberg's constant = 8.314 KJ/K
M = molar mass of water = 18 g/ mol
Now, using eqn (1):
[tex]V = \frac{m\bar{R}T}{P}[/tex]
[tex]V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}[/tex]
[tex]V = 2762.44 m^{3}[/tex]
Therefore, the volume of the tank is [tex]V = 2762.44 m^{3}[/tex]
b) After extracting 100 kg of water, amount of water left, m' = m - 100
m' = 2000 - 100 = 1900 kg
The remaining water reaches thermal equilibrium with surrounding temperature at T' = [tex]30^{\circ}C[/tex] = 303 K
At equilibrium, volume remain same
So,
P'V = m'[tex]\bar{R}[/tex]T'
[tex]P' = \frac{1900\times \frac{8.314}{18}\times 303}{2762.44}[/tex]
Therefore, the final pressure is P' = 96.258 kPa