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:
c
Explanation:
if someone is wrong that they can help with
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
Answer:
i want coins sorry use a calculator or sum
Explanation:
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Answer:
Explanation:
The schedule using shortest remaining time, non-preemptive priority and round Robin with quantum number 30 is shown in the attached file, please kindly go through it to access the answer.
