Answer:
The work transfer per unit mass is approximately 149.89 kJ
The heat transfer for an adiabatic process = 0
Explanation:
The given information are;
P₁ = 1 atm
T₁ = 70°F = 294.2611 F
P₂ = 5 atm
γ = 1.5
Therefore, we have for adiabatic system under compression
Therefore, we have;
The p·dV work is given as follows;
Therefore, we have;
Taking air as a diatomic gas, we have;
The molar mass of air = 28.97 g/mol
Therefore, we have
The work done per unit mass of gas is therefore;
The work transfer per unit mass ≈ 149.89 kJ
The heat transfer for an adiabatic process = 0.
Answer:
Check the explanation
Explanation:
The loop invariant has to satisfy some amount of requirements to be of good use. Another complex factor as to why a loop is the question of loop termination. A loop that doesn’t terminate can’t invariably be correct, and in fact the computation in whatever form amounts to nothing. The total axiomatic description of a while construct will have to involve all of the following to be true, in which I is the loop invariant:
P => I
{I and B} S {I}
(I and (not B)) => Q
Then the loop terminates
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
Answer:
Hello your question is incomplete attached below is the missing part and answer
options :
Effect A
Effect B
Effect C
Effect D
Effect AB
Effect AC
Effect AD
Effect BC
Effect BD
Effect CD
Answer :
A = significant
B = significant
C = Non-significant
D = Non-significant
AB = Non-significant
AC = significant
AD = Non-significant
BC = Non-significant
BD = Non-significant
CD = Non-significant
Explanation:
The dependent variable here is Time
Effect of A = significant
Effect of B = significant
Effect of C = Non-significant
Effect of D = Non-significant
Effect of AB = Non-significant
Effect of AC = significant
Effect of AD = Non-significant
Effect of BC = Non-significant
Effect of BD = Non-significant
Effect of CD = Non-significant
