19 THE FIRST LAW OF THERMODYNAMICS 19 1 a IDENTIFY and SET UP The pressure is constant and the volume increases The pV diagram is sketched in Figure 19 1 Figure 19 1 V 2 V 1 b W p dV V 2 V 1 Since p is constant W p dV p V 2 V 1 19 2 19 3 T 1 p V 2 2 nRT 1 nRT 2 p V so 1 1 subtracting the two equations gives The problem gives T rather than p and V so use the ideal gas law to rewrite the expression for W EXECUTE pV nRT V nR T T p V 1 2 1 2 W nR T is an alternative expression for the work in a constant pressure process for an Thus 2 ideal gas Then 2 00 mol 8 3145 J mol K 107 C 27 C EVALUATE The gas expands when heated and does positive work IDENTIFY At constant pressure so V SET UP is positive which means that T must also be positive W p V nR T W nR T 2 1330 J Since the gas is doing work it must be expanding T 1 has the same numerical value in kelvins and in C 3 T K T C and 0W the gas expands When p is constant and V increases T increases pV nRT W nRT ln p p 1 2 says V R 48 1 K EXECUTE T 2 40 10 J 6 mol 8 3145 J mol K 75 1 C 8 3145 J mol K WT nR T 27 0 C 48 1C 2 EVALUATE When IDENTIFY Example 19 1 shows that for an isothermal process decreases when p increases and T is constant SET UP p 2 EXECUTE a The pV diagram is sketched in Figure 19 3 2 00 mol 8 314 J mol K 338 15 K ln 65 0 273 15 338 15 K p 13 T b W p 1 p 3 1 6180 J EVALUATE Since V decreases W is negative Copyright 2012 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 19 1 19 2 Chapter 19 Figure 19 5 19 4 p gauge p air and T is constant the maximum number of moles of air in the lungs is when pV is a In calculating the enclosed area only changes in 3 p 3 1 L 10 m Figure 19 3 IDENTIFY The work done in a cycle is the area enclosed by the cycle in a pV diagram SET UP a 1 mm of Hg 133 3 Pa pressure enter and you can use gauge pressure b Since pV nRT maximum In the ideal gas law the absolute pressure p air EXECUTE a By counting squares and noting that the area of 1 square is 1 mm of Hg 0 1 L we estimate that the area enclosed by the cycle is about 7 5 mm of Hg L 1 00 N m positive b The maximum pV is when V 1 028 10 Pa pV nRT 3 1 4 L 1 4 10 m 1 mm of Hg 1 torr 760 torr pV 760 mm of Hg must be used and so 144 N m 771 torr p gauge 11 torr p p air 3 p 5 The net work done is pV RT n max max The maximum pV is 144 N m max 0 059 mol 8 315 J mol K 293 K 19 5 EVALUATE While inhaling the gas does positive work on the lungs but while exhaling the lungs do work on the gas so the net work is positive IDENTIFY Example 19 1 shows that for an isothermal process SET UP For a compression V decreases W is negative so p p ln 1 2 T 295 15 K W nRT W 468 J Solve for 1 p EXECUTE a W nRT p 1 p 2 p 1 ln p 2 468 J W nRT e p e 2 0 6253 0 6253 0 305 mol 8 314 J mol K 295 15 K W nRT 0 942 atm W nRT p 1 b In the process the pressure increases and the volume decreases The pV diagram is sketched in Figure 19 5 EVALUATE W is the work done by the gas so when the surroundings do work on the gas W is negative The gas was compressed at constant temperature so its pressure must have increased which means that p1 p2 which is what we found 1 76 atm e Copyright 2012 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 19 6 a IDENTIFY and SET UP The pV diagram is sketched in Figure 19 6 The First Law of Thermodynamics 19 3 19 7 19 8 19 9 4 3 5 4 tot W so 3 3 2 0V 0W W p V W 1 2 2 4 00 10 J 2 4 00 10 J W is negative since 3 5 00 10 Pa 0 120 m 0 200 m Figure 19 6 b Calculate W for each process using the expression for W that applies to the specific type of process EXECUTE 1 2 p is constant so the volume decreases in the process W EVALUATE The volume decreases so the total work done is negative IDENTIFY Calculate W for each step using the appropriate expression for each type of process SET UP When p is constant W p V When W EXECUTE a 0 24 p V system is 2 2 b For the process in reverse the pressures are the same but the volume changes are all the negatives of those found in part a so the total work is negative of the work found in part a EVALUATE When IDENTIFY The gas is undergoing an isobaric compression so its temperature and internal energy must be decreasing SET UP The pV diagram shows that in the process the volume decreases while the pressure is constant 1 L 10 m 1 which is the area in the pV plane enclosed by the loop W W W W V 0 p V V 2 2 V 1 The total work done by the 0 V W 32 1 p 1 1 atm 1 013 10 Pa 0W W and 0W and when p V 2 1 0W and V V W 13 0 0 13 24 41 41 32 5 3 3 EXECUTE a pV nRT n R and p are constant so constant V T nR p V a T a V b T b 1 50 atm 0 125 L 0 500 L 57 0 J and W is negative since the volume decreases V b V a T b T a 0 500 L …