A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank (m3 ). Later, a pump is used to extract 100 kg of water from the tank. The water remaining in the tank eventually reaches thermal equilibrium with the surroundings at 30 °C). Determine: b) the final pressure (kPa).

Respuesta :

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

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