A 3.5-m3 rigid tank initially contains air whose density is 2 kg/m3 . The tank is connected to a high-pressure supply line throu
1 answer:
Answer:
Explanation:
First, we find the mass of the air originally in the tank.
Density is given as mass divided by volume. It is given as:
Therefore, mass is:
Density of air = ; Volume of the tank =
The mass of the air initially in the tank is 7 kg.
After air is allowed to enter, the mass changes.
New density =
The new mass will be:
We can now find the mass of air that has entered the tank:
Mass of air that entered tank = New mass of air - Original mass of air
M = 22.75 - 7.0 = 15.75 kg
The mass of air that entered the tank is 15.75 kg.
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Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-
a=
b=
from virial equation
And
a=
b=
Now calculate V1 and V2 at given condition
Substitute given values = 1 x 10^5 , T = 373.15 and given values of coefficients we get
Solve for V1 by iterative or alternative cubic equation solver we get
Similarly solve for state 2 at P2 = 50 bar we get
Now
a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is
a=
b=
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series