Engine / thermo question?

Question

A heat engine using 140 of helium as the working substance follows the cycle shown in the figure.

Figure has a "triangle" with coordinates: 1=(1000, 1), 2=(1000, 5), 3=(Vmax, 1)
The cycle starts with 1 and ends at 3. Segment 2-3 is an isotherm and has a curve like half an upward-facing parabola.

What is the temp of the gas at point 1? point 2? point 3?
What is the thermal efficiency?
What is the maximum possible efficiency of a heat engine that operates between and Tmax and Tmin?

Answer 1

To start: 140grams of He is just 35Moles

ANSWER A
At Point 1
using the Universal Gas Law: pV=nRT
1 * 1000 = 35 * 8.3144 * T
so T = 3.436 Kelvins

At Point 2
using the Universal Gas Law: pV=nRT
5 * 1000 = 35 * 8.3144 * T
so T = 17.182 Kelvins

At Point 3
The process is Isothermal
the temperature does not change
so T = 17.182 Kelvins

Now on your triangular pV diagram draw the Mechanical Work in and out arrows crossing the horizontal and curved sides
and draw the heat in and out arrows crossing the vertical and horizontal sides.

ANSWER B
The efficiency of the engine is z = NetWorkOut/HeatIn

NetWork is work done expanding from Point2 to Point3
minus work done in compressing from Point3 to Point1
WorkOut = nRT*Ln(V3/V2) = 35*8.3144*17.182*Ln5
WorkOut= 8047Joules
WorkIn at constant pressure = PV
= 1*(5000-1000) = 4000Joules
Net Work = WorkOut - WorkIn = 4047Joules

Heat In = mC(v)*diffTemp = 0.14 * 5200 *13.746 = 10007J

so z = 4047/10007 = 0.4

ANSWER C
The maximum efficiency between the heat source and sink is: z = 1 - TL/TH
z = 1 - 3.436/17.182 = 0.8

So we can see that as we have not followed the ideal adiabatic/ isothermal cycle then irreversibility has cost us half our efficiency.

NOTE: If this question was in the Finals Paper of a Phys cs Degree Course you would be expected to do the analysis of the adiabatic/ isothermal cycle and derive the above formula - not just assume it.

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