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$\bullet$$\bullet \mathrm{A}$ volume of $\operatorname{air}$ (assumed to be an idealgas) is first cooled with- out changing its volume and then expanded with- out changing its pressure, as shown by the path abc in Fig. 15.38 (a) How does the final temperature of the gas compare with its initial temperature? (b) How much heat does the air exchange with its surroundings during the process $a b c ?$ Does the air absorb heat or release heat during this process? Explain. (c) If the air instead expands from state $a$ to state $c$ by the straight-line path shown, how much heat does it exchange with its surroundings?

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Physics 101 Mechanics

Chapter 15

Thermal Properties of Matter

Temperature and Heat

The First Law of Thermodynamics

The Second Law of Thermodynamics

University of Washington

Simon Fraser University

University of Winnipeg

McMaster University

Lectures

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Air (a diatomic ideal gas)…

Here an ordeal guesses first cooled from a to be at constant volume, then, from b to c the arden gets the expanded air constant pressure. Now we want to compare the temperature air state, air and air state temperature at state is t a and the temperature air state c is t c so by using a real gas equation, we have p v is equal to n r t from here. We get t is equal to p v upon and are for sited. Have our relation is temperature at a is equal to pressure at a into volume at a upon n and are here and are are constants foster fostat b we have temperature sorry for state. We have temperature at c is equal to pressure at c times. Volume c upon n times are here number of moles and value of r is constant acetate and state b. Now, by dividing temperature at c and temperature at we get t c upon, th air is equal to p c v c upon n r, divided by pressure 8 a times volume 8 a upon. Nor so we get temperature at c upon temperature at a is equal to pressure c times, volume at c divide by pressure 8 a times volume at a so we get temperature at c upon temperature at a is equal to here. Pressure at c from the graph is 1 multiply, 10 to the power 5 pascal and volume 8 c is 0.06 cubic meter upon pressure at a is 3 multiply, 10 to the power 5 pascal and volume at a is 0.02 cubic meter. So we get t c upon t a is equal to 1, which implies that temperature at c is equal to temperature 8, a as a final temperature is equal to initial temperature. Now, in the second part of this problem, we want to calculate the amount of heat released or absorbed in the bath a b c, so he released or absorbed in part. A b c can be expressed, as sum of heat released, are absorbed in part, a b plus heat in the path. Let this name this equation now we will first calculate the heat release or absorbed in a bath a b. Then we will calculate the heat released or absorbed in part b c now by using forsle of the termodanomics for part, a b we have a b is equal to work done from a to b, plus changing internal energy, from a to be as there is no Check in ter no check in volume from a to b, so work done from a to b is equal to 0, so we have heat released or absorbed from a to b is equal to change in internal energy from a to b. Now far apart, b c, we have c from b c is equal to work done from b to c plus change in into energy. From b to c o q from b to c is equal to here. Work done from b to c is equal to pressure. At b, into changing volume, here, final volume is v c, and initial volume is v b, plus changing internal is delta. U b c: this is heat released or absorbed in part b c now, by putting q a b and c b c in equation 1. So we get equation 1 coin part a b c is equal to a b which is equal to delta. U a b, plus c b c that is equal to check in internal energy, from b to c plus pressure at b into v c minus v b. Now we have in part a b c is equal to the first 2 terms is sum of 2 internal energy changes here, the internal energy change from at a to b, plus internal energy change, from b to c that is equal to internal energy. From a to c, plus, b, b into v c minus v b, a temperature of point, a is equal to temperature of point c, as we have calculated already in part a so change in internety from point a and c will be equal to 0. So we get q in the path. A b c is equal to p b into v c minus v b. Now, by putting different values, we get q in the path. A b c is equal to her. Pressure at b is 1.0 multiply, 10 to the power 5 pascal and volume at c is 0.06. Minus volume and b is 0.02 cubic meter. So we get in the path. A b c is equal to 4 kilo. Joule, here heat is in positive to show that the system absorbs heat from the surrounding now. In the c part, we want to calculate the heat when the gas expands from a to c so again by using fossil ap thermodynamics. We have at first after anemic. We have come from in the path. A to c is equal to work done from a to c plus. Changing internal energy in the path is temperature of c is equal to temperature. Therefore, change in internally from a to c is equal to 0. So we get q in the path. A c is equal to work done in the path, a c and the work done in pressure versus volume graph is equal to area under the graph here. The area under the part, a c is a trapezium having parallel sides. Equal to b c and p a in length and the distance between the sides is v c minus, where his area of trapezium is equal to 1. Upon 2, some of peel sides multiply perpendicular distance between the peer sides, so we get area equal to 1. Upon 2, here some of a side is p, a plus p c, and the perpendicular distance between the placide is v c minus v, a t, so we get heat, released or absorbed in the part. A c is equal to 1 upon 2 into here p. A pressure at a is 3 multiplied 10 to the power 5. Pascal plus pressure at c is 1 multiplied. 10 to the power 5 pascal into volume is 0.06 minus volume at a is 0.02 cubic meter. So by computing all these values we get heat release are absorbed in part, a c equal to at kilo. Joules, here heat is in positive sine. This shows that the system absorbs heat from surrounding in the path.

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