A mass of 5 kg of saturated liquid-vapor mixture of water is contained in a piston-cylinder device at 125 kPa. Initially, 2 kg o
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Answer:
Relative density = 0.545
Degree of saturation = 24.77%
Explanation:
Data provided in the question:
Water content, w = 5%
Bulk unit weight = 18.0 kN/m³
Void ratio in the densest state, = 0.51
Void ratio in the loosest state, = 0.87
Now,
Dry density,
= 17.14 kN/m³
Also,
here, G = Specific gravity = 2.7 for sand
or
e = 0.545
Relative density =
=
= 0.902
Also,
Se = wG
here,
S is the degree of saturation
therefore,
S(0.545) = (0.05)()2.7
or
S = 0.2477
or
S = 0.2477 × 100% = 24.77%
Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.
Explanation: <u>Dalton's</u> <u>Law</u> <u>of</u> <u>Partial</u> <u>Pressure</u> states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:
The proportion of each individual gas in the total pressure is expressed in terms of <u>mole</u> <u>fraction</u>:
= moles of a gas / total number moles of gas
The rigid tank has total pressure of 1MPa.
- Nitrogen gas:
molar mass = 14g/mol
mass in the tank = 2000g
number of moles in the tank: = 142.85mols
- Carbon Dioxide:
molar mass = 44g/mol
mass in the tank = 4000g
number of moles in the tank: = 90.91mols
Total number of moles: 142.85 + 90.91 = 233.76 mols
To calculate partial pressure:
For Nitrogen gas:
= 0.6
For Carbon Dioxide:
0.4
Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.