Question

An ideal gas with 7 degrees of freedom begins at a pressure of 8.5 atm, temperature of 25 degrees Celsius and volume of 120L.

A) How many moles of gas are there?

B) The gas expands isobarically to 200L. What is the new temperature?

C) How much work was done in the expansion?

D) By how much did the internal energy increase?

E) What amount of heat flowed into the gas?

G) The gas then depressurizes isochorically to the original temperature. How much heat flowed out of the gas?

H) Finally the gas contracts isothermally back to its original volume.

I) How much work was done on the gas in the Sketch this thermal cycle on a PV diagram. contraction?

J) How much heat flowed out of the system?

K) What is the efficiency of this thermal cycle?

L) What is the Carnot efficiency for an ideal heat engine running between the same high and low operating temperatures?

Answer 1

Initial condition: P₁ = 8.5 atm ; V₁ = 120 L ; T₁ = 25+273 = 298 K degree of freedom (f) = 7

therefore c_v = f/2*R = 7/2*R and c_p = c_v + R = 9/2*R

(A) moles of gas (n) = P₁*V₁/R*T₁ = 8.5*120/0.082*298 = 41.74

(B) For the given isobaric process we get V₂ = 200 L ; P₂ = P₁ = 8.5 atm

hence using gas law P₁*V₁/T₁ = P₂*V₂/T₂ or T₂ = 8.5*200*298/8.5*120 = 496.67 K

(C) Work done in expansion (ΔW) = P₁*(V₂-V₁) = 8.5*10⁵*(200-120)*10^-3 = 68000 J

(D) Increase in internal energy (ΔU) = n*c_v*(T₂ - T₁) = 41.74*3.5*8.314*(496.67-298) = 241303 J

(E) Amount of heat flowed into the gas (ΔQ₁) = ΔU + ΔW (using first law of thermodynamics)

Hence ΔQ₁  = 241303 + 68000 = 309303 J

(F) Now for the given Isochoric process: V₃ = V₂ = 200 L ; T₃ = 298 K

Hence heat given to the system (ΔQ₂) = n*c_v*(T₃ - T₂) = 41.74*3.5*8.314*198.67 = 241303 J

(H) In the final isothermal process : T₄ = T₁ = 298 K ; V₄ = V₁ = 120 L

So work done ( ΔW) = n*R*T₁*ln(V₁/V₃) = 41.74*8.314*298*ln(120/200) = -52826.4 J

(I) As the change in internal energy is zero in isothermal process therefore

heat flown out of the system (ΔQ₃) = -52826.4 J

(J) Net heat given to system (ΔQ_in) = ΔQ₁ + ΔQ₂ = 309303 + 241303 = 550606 J

also total amount of work done by the system (ΔW_tot) = 68000 - 52826.4 = 15173.6 J

Therefore the efficiency of the engine (η) = ΔW_tot/ΔQ_in = 15173.6/550606 = 0.027

(K) Efficiency of Carnot engine operating between 298 K and 496.67 K = 1 - 298/496.6 = 0.399