A circuit has a source voltage of 15V and two resistors in series with a total resistance of 4000Ω .If RI has a potential drop o
1 answer:
Answer:
1500Ω
Explanation:
Given data
voltage = 15 V
total Resistance = 4000Ω
potential drop V = 9.375 V
To find out
R2
Solution
we know R1 +R2 = 4000Ω
So we use here Ohm's law to find out current I
current = voltage / total resistance
I = 15 / 4000 = 3.75 × A
Now we apply Kirchhoffs Voltage Law for find out R2
R2 = ( 15 - V ) / current
R2 = ( 15 - 9.375 ) / 3.75 ×
R2 = 1500Ω
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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