Answered: A heat engine takes an ideal monoatomic…

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter2: Kinematics

Section: Chapter Questions

Problem 19CQ: What is the last thing you should do when solving a problem? Explain.

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Question

A heat engine takes an ideal monoatomic gas through the cycle in this pV diagram. Step ab is isochoric, bc is isothermal, and ca is isobaric. At point a, the pressure is 2.00 atm, the volume is 2.00 L, and the temperature is 300 K. At point b, the temperature is 600K.
a) Calculate the heat added or removed, change in internal energy, and work done for each step and complete the table below.

b) What is the efficiency of this engine? 1 atm = 1.01×105 Pa

а)
Q(J)
AU (J)
W (I)
a → b
b →c
с) efficiency -

Expert Solution

Step 1

The first law of thermodynamics states that:

$∆ U = Q - W w h e r e ∆ U i s t h e c h a n g e i n i n t e r n a l e n e r g y o f t h e c l o s e d s y s t e m Q d e n o t e s t h e q u a n t i t y o f h e a t e n e r g y s u p p l i e d t o t h e s y s t e m W d e n o t e s t h e w o r k d o n e b y t h e s y s t e m$

a)	Q(J)	$∆ U (J) = n \times \frac{3}{2} \times R \times ∆ T$	W(J)
a $\to$ b	22597.82	22597.82	0
b $\to$ c	9072	0	9072
c $\to$ a	-23001.87	-22597.82	-404

At point a, P= 2 atm = 2.02 X 10⁵ Pascal, V= 2L and T= 300 K

Using the gas equation $P V = n R T \equiv n = \frac{R T}{P V} = \frac{8.314 \times 300}{202000 X 0.002} = 6.174 m o l e s$

For a monoatomic gas $∆ U = n \times \frac{3}{2} \times R \times ∆ T$

Pressure at b = $\frac{P_{a} T_{b}}{T a} = \frac{2 \times 600}{300} = 4 a t m$ [a $\to$ b process]

Volume at c= $V c = \frac{V_{b} P_{b}}{P_{c}} = \frac{2 \times 4}{2} = 4 l i t e r s$

Work Done in a Isothermal process = $W = n R T \times \log [\frac{V_{f}}{V_{1}}] = 6.174 \times 600 \times 8.134 \times \log (\frac{4}{2}) = 9072 J$

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