A heat engine does work by using a gas at an initial pressure of 1000 Pa and vol

Question

A heat engine does work by using a gas at an initial pressure of 1000 Pa and volume .1m 3. Step-by-step, it then increases the pressure to 10,000 Pa (at constant volume), increases the volume to .15m3 (at constant pressure), decreases the pressure back to 1,000 Pa (at constant volume) and returns the volume back to .1m3 (at constant pressure).

G. How much heat is added to the gas in the first step? (0 J, 450 J, 500 J, 750 J, or 1350 J)

H. How much heat is added to the gas in the second step? (0 J, 450 J, 500 J, 750 J, or 1350 J)

I. How much heat is added to the gas in one complete cycle? (0 J, 450 J, 500 J, 750 J, or 1350 J)

I;ve figured that A=0 J, B = 500 J, and C = 450 J but I can't seem to work out D - I to make enough sense

Explanation / Answer

(1)
Work done BY the gas is given by the integral
W = ∫ p dV from initial to final volume

In the isochoric steps of the process no work is done, cause volume does no change, i.e
W₂ = W₃ = 0

For an ideal gas undergoing an isothermal change of state:
p∙V = n∙R∙T = constant
Hence:
p∙V = p_initial ∙ V_initial
<=>
p = p_initial ∙ V_initial/V

So the work done in thermal process is
W = p_initial∙V_initial ∫ 1/V dV from initial to final volume
= p_initial∙V_initial∙ ln(V_final/V_initial)

Using ideal gas law, you can substitute p and V:
p_initial∙V_initial = n∙R∙T_initial

Hence:
W = n∙R∙T_initial ∙ ln(V_final/V_initial)

For the expansion process.
Wâ‚ = 0.5mol ∙ 8.3145J/molK ∙ 800K ∙ ln(6000cm ³/1000cm ³)
= 5960J

Work done by the gas in compression process is
W₃ = 0.5mol ∙ 8.3145J/molK ∙ 400K ∙ ln(1000cm ³/1000cm ³)
= -2980J

So the net work done by the engine per cycle is:
W = W₠+ W₂ + W₃ + W₄
= 5960J + 0J -2960J + 0J
= 2960J


2)
To calculate efficiency you need the heat transferred to the engine.
You find it from internal energy considerations
Change of internal energy of an ideal gas is given by:
∆U = n∙Cv∙∆T
For a monatomic gas
Cv = (3/2)∙R

On the other hand change of internal energy equals heat transferred to the gas minus work done By it (or plus work done on it):
∆U = Q - W

For the isothermal steps
∆U = Q - W = 0
=>
Q = W

For the isochoric steps
W = 0
=>
Q = ∆U = (3/2)∙n∙R∙∆T

Hence:
Qâ‚ = Wâ‚ = 5960J
Q₂ = (3/2)∙0.5mol∙8.3145J/mol ∙ (400K - 800K) = -2494J
Q₃ = W₃ = -2980J
Q₄ = (3/2)∙0.5mol∙8.3145J/mol ∙ (800K - 400K) = +2494J

So the total heat transferred to the engine is
Q_in = Qâ‚+ Qâ‚„ = 5960J + 2494J = 8454J

The total heat rejected by the engine is
Q_out = - Q₂ - Q₃ = 2494J + 2980J = 5474J

It efficiency is the ratio of net work done to heat absorbed:
η = W / Q_in
= 2980J / 8454J
= 0.352
= 35.2%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.