A vessel contains 10^-3m^3 of helium gas at 3K and 10^3 Pa. Take the zero of int

Question

A vessel contains 10^-3m^3 of helium gas at 3K and 10^3 Pa. Take the zero of internal energy of helium to be at this state.

(a) The temparature is raised at constant volume to 300 K . Assuming helium to behave like an ideal monatomic gas, how much heat is absorbed, and what is the internal energy of the helium? Can this energy be regarded as the result of heating or working

(b) The helium is now expanded adiabatically to 3K. How much work is done, and what is the new internal energy? Has heat been converted to work without compesation, thus violating the second law

(c ) The helium is now compressed isothermally to its original volume. What are the quantities of heat and work in this process? What is the thermal efficiency of the cycle on a PV diagram

Explanation / Answer

V1 = 10^-3 m^3 ; T1 = 3 K , P1 = 10^3 Pa

U1 =0

a)

T2 = 300 K , V2 = 10^-3 m^3   P2 = unknown

so, by ideal gas law,

P1*V1 = n*R*T1

so, (10^3 * 10^-3) = n*8.314*3

n = 1/ ( 8.314*3 ) = 0.040093 moles

P1*V1 / T1= P2*V2/T2

so, 10^3 * 10^-3 / 3 = P2 * 10^-3 / 300

P2 = 10^5 Pa

Cv = 1.5*R = 1.5*8.314 = 12.471 J / (mol .K)

In constant volume process, Internal energy rise , U2 - U1 = n*Cv*(T2 -T1) = 0.040093* 12.471* (300 - 3) = 148.499941 J

so , heat supplied = Increase in internal energy = 148.499941491 J

internal energy of the helium = 148.499941491 J

this energy be regarded as the result of heating  

work done = 0 in this case.

b)

Now, T3 = 3 K ,

For adiabatic process

T3/T2 = (V3/V2) ^ (g-1)

g = Cp/Cv = 5/3 for monoatomic gas

so, 3/300 = ( V3/ 10^-3) ^ ( 5/3 -1)

V3 = 10^-6 m^3

In adiabatic expansion

work done = -n*Cv*(T3 - T2) = -12.471 * 0.040093 * ( 3 - 300) = 148.499941491 J

new internal energy = U3 = U2 - W = 148.499941491 - 148.499941491 = 0 J

new internal energy = 0 J

heat has been converted to work without compesation, But you have to observe that volume(10^-6) is not the same as initial volume (10^-3 m^3) , so you cannot say that it is violating the second law.

c)

We have v4 = v1 = 10^-3 m^3 ; T4 = 3 K

In isothermal process, work done by external = heat absorbed = nRT ln ( v4/v3) = 0.040093*8.314*3* ln ( 10^-6 / 10^-3)

work done by external = -6.907752 J ,

work is done by the gas = 6.907752 J ,

so heat is absorbed by the gas. heat absorbed by the gas. = 6.907752 J ,

thermal efficiency of the cycle on a PV diagram =

Total heat absorbed = 6.907752 + 148.499941491 = 155.407693491 J

Total work done = 148.499941491 J

SO thermal efficiency = work done / Heat given = 148.499941491 / 155.407693491 = 0.95555

Thermal efficiency = 0.95555 , or 95.555 %

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