Given:
mass of water, m = 2000 kg
temperature, T =
= 303 K
extacted mass of water = 100 kg
Atmospheric pressure, P = 101.325 kPa
Solution:
a) Using Ideal gas equation:
PV = m
T (1)
where,
V = volume
m = mass of water
P = atmospheric pressure
![\bar{R} = \frac{R}{M}](https://tex.z-dn.net/?f=%5Cbar%7BR%7D%20%3D%20%5Cfrac%7BR%7D%7BM%7D%20)
R= Rydberg's constant = 8.314 KJ/K
M = molar mass of water = 18 g/ mol
Now, using eqn (1):
![V = \frac{m\bar{R}T}{P}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bm%5Cbar%7BR%7DT%7D%7BP%7D)
![V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B2000%5Ctimes%20%5Cfrac%7B8.314%7D%7B18%7D%5Ctimes%20303%7D%7B101.325%7D)
![V = 2762.44 m^{3}](https://tex.z-dn.net/?f=V%20%3D%202762.44%20m%5E%7B3%7D)
Therefore, the volume of the tank is ![V = 2762.44 m^{3}](https://tex.z-dn.net/?f=V%20%3D%202762.44%20m%5E%7B3%7D)
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' =
= 303 K
At equilibrium, volume remain same
So,
P'V = m'
T'
Therefore, the final pressure is P' = 96.258 kPa