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Get Full Access to University Physics - 13 Edition - Chapter 20 - Problem 40p
Get Full Access to University Physics - 13 Edition - Chapter 20 - Problem 40p

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# A heat engine takes 0.350 mol of a diatomic ideal gas

ISBN: 9780321675460 31

## Solution for problem 40P Chapter 20

University Physics | 13th Edition

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University Physics | 13th Edition

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Problem 40P

A heat engine takes 0.350 mol of a diatomic ideal gas around the cycle shown in the p?V-diagram of ?Fig. P20.36. Process 1 ? 2 is at constant volume, process 2 ? 3is adiabatic and process 3 ? 1 constant pressure of 1.00 atm. The value of ? for this gas is 1.40. (a) Find the pressure and volume at points 1, 2, and 3. (b) Calculate ?Q, W?, and ?U for each of the three processes. (c) Find the net work done by the gas in the cycle. (d) Find the net heat flow into the engine in one cycle. (e) What is the thermal efficiency of the engine? How does this compare to the efficiency of a Carnot-cycle engine operating be-tween the same minimum and maximum temperatures T? a1? T? ? 2?

Step-by-Step Solution:

Solution 40P Step 1: Considering the pv diagram Find pressure and volume at points 1 2 and 3 Data given P 1 1.0 atm = 1.013 × 10 5 T = 300K 1 T = 600 K 2 T3= 492 K n = 0.350 mol Case (1) To find volume V at point 1 1 We shall consider the equation PV = nRT Rearrange to find V we get V = nRT1 1 P1 Substituting we get V = 0.350 ×8.312×300 K 1 1.013×10 3 V 1 8.62 × 10 m 3 Hence at point 1 pressure is P = 1.0 atm = 1.013 × 10 and 5 volume is 1 V = 8.62 × 10 m 3 3 1 Case (2) From the graph we can note that volume at point 2 is same as point 1 Hence we get V = V 1 2 Pressure at point 2 We can obtain using P 1T =1P /T 2 2 Rearranging to find P 2 P T /T = P 1 2 1 2 Substituting we get 5 ( 1.013 × 10 × 600 K) / 300K2 = P 2 P 2 2.02 × 10 or 2.0 atm Hence at point 2 pressure is P = 2.0 atm = 2.02 × 10 and 5 volume is 2 3 3 V =28.62 × 10 m Case (3) To find volume at point 3 This can be obtained using the relation V /T = V /T 1 1 3 3 Rearranging to find V w3 get ( V T )/T = V 1 3 1 3 Substituting the values we get 3 3 ( 8.62 × 10 m × 492 K)/300K = V 3 3 V 3 1.41 × 10 m 3 From the graph we can note that pressure at point 1 = point 3 Hence we have P = P 1 3 5 Hence at point 1 pressure is P 1 1.0 atm = 1.013 × 10 and volume is 3 3 V 3 1.41 × 10 m

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##### ISBN: 9780321675460

This full solution covers the following key subjects: cycle, engine, process, Find, gas. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 40P from chapter: 20 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. Since the solution to 40P from 20 chapter was answered, more than 1113 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “A heat engine takes 0.350 mol of a diatomic ideal gas around the cycle shown in the p?V-diagram of ?Fig. P20.36. Process 1 ? 2 is at constant volume, process 2 ? 3is adiabatic and process 3 ? 1 constant pressure of 1.00 atm. The value of ? for this gas is 1.40. (a) Find the pressure and volume at points 1, 2, and 3. (b) Calculate ?Q, W?, and ?U for each of the three processes. (c) Find the net work done by the gas in the cycle. (d) Find the net heat flow into the engine in one cycle. (e) What is the thermal efficiency of the engine? How does this compare to the efficiency of a Carnot-cycle engine operating be-tween the same minimum and maximum temperatures T? a1? T? ? 2?” is broken down into a number of easy to follow steps, and 134 words. University Physics was written by and is associated to the ISBN: 9780321675460.

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