Is the refractive index constant?
2 answers:
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Answer:
The answers can be found by considering the isothermal expansion equation as well as the ideal gas equation from where we have
The temperature T = 602.64K and the final pressure P = 110.24MPa
Explanation:
Numbeer of moles of gas = 3.00 mol
initial volume = 230.8×10−6 m3
final volume = 133.4×10−6 m3 .
released energy = 8240 J
Temperature = Constant = T
Pressure = =unknown
From the relation the combined ideal gas law, PV = nRT
Where R = 8.314 4621.JK−1mol−1
we have The release energy from compression P1V1 -P2V2
-qrev = -nRTln() = 8240J
n = 3
Hence -nRTln() = 3×8.314 462×ln
× T= -8240 J
or -13.67×T = -8240J, thus T = -8240/-13.67 = 602.64K
The Final pressure is given by
PV = n×R×T from where we have V = final volume thus
P = (n×R×T)/V = (3×8.134×602.64)÷(133.4×) = 110237041.1 N/
= 110.237MPa
The strength of the electric field will be 215760 NC⁻¹. The concept of the electric field strength is used in the problem.
<h3>What is electric file strength?</h3>The electric field strength is defined as the ratio of electric force and charge. Its unit is NC⁻¹.
The given data in the problem is;
E is the Electric Field Strength
k is the Colomb's constant = 8.99 x 10⁹ N.m²/C²
q is the magnitude of charge = 6 μC = 6 x 10⁻⁶ C
r is the distance = 0.5 m
The electric field strength is given by the formula;
Hence the strength of the electric field will be 215760 NC⁻¹.
To learn more about the electric field strength refer to the link;