Answered: A heat engine takes an ideal monoatomic…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

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

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions