1 Kmol of N(2) is confined in a piston initally at 300k with a volume of .1(m^3)

Question

1 Kmol of N(2) is confined in a piston initally at 300k with a volume of .1(m^3). The piston is well insulated. A microwave gun then heats the gas for 30 seconds, where the power of the microwave is 10MW and the absorbtion coeffecient of the gas is = .1 During the microwave heating the N(2) gas expands at constant pressure. Use tables A-2, A-1, A-18 for Cp, Cv, Rgas, etc...

a) What is the initial pressure?

b) What is the temp. of the gas assuming the heat capacity is constant?

c) what is the final temp. from the change in enthalpy?

d) what is the volume of the gas after heating?

e) what is the change in entropy using the 1st TdS relation? Assuming Cv(T)= Cv(@T1)

f) What is the change in entropy using the 2nd TdS relation? Assuming Cp(T)= Cp(@T1)

g)What is the change in entropy using the 2nd TdS relation? Assuming Cp(T)= [Cp(@T1) + Cp(@T2)]/2

h) What is the change in entropy using the 2nd TdS relation? Using S2 and S1

i) What is the work done by the gas?

Explanation / Answer

n = 1000 mol

Cp =1.04 KJ/Kg K

Cv = 0.743 KJ/Kg K

Qp = 0.1*10*10^6*30 = 30 MJ/Kg

i for initial

Vi = 0.1 m^3

Ti = 300 K

a)Pi*Vi = n*R*Ti

Pi*0.1 = 1000*8.314*300

Pi = 24.94 MPa

b) and c) For Constant Pressure process both heat capacity and enthalpy are same(?H = Qp)

Qp = Cp(T2-T1)

30*10^6 = 1.04*10^3(T2-300)

T2 = 29,146 K for both cases

d)From Ideal gas Equation for const Pressure

V2 = V1*(T2/T1) = 9.72 m^3

e) dU = Cv(T2-T1)= 21.45 MJ/Kg

dS = dU/T1 = 0.0715 MJ/Kg K

f) dS = dQ/T1 = 0.1 MJ/Kg K

g)T = 14723

dS = dQ/T = 2.03 KJ/Kg K

h)W = P(V2-V1) = 239.92 MJ

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