HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS

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1 HET, WORK, ND THE FIRST LW OF THERMODYNMICS 8 EXERCISES Section 8. The First Law of Thermodynamics 5. INTERPRET We identify the system as the water in the insulated container. The problem involves calculating the work done to raise the temperature of a system, so the first law of thermodynamics comes into play. DEELOP Because the container is a perfect thermal insulator, no heat enters or leaves the water inside of it. Thus, Q and the first law of thermodynamics in Equation 8. gives Δ U Q+ W W, where W is the work done by shaking the container. The change in the internal energy of the water is determined from its temperature rise, Δ U mc Δ T (see comments in Section 6. on internal energy). Combining these two expressions for the internal energy change allows us to find the work done. ELUTE The work done on the water is W Δ U mcδt (. kg) 4.84 kj/ ( kg K) ( 7. K) 9. kj SSESS ccording to the convention adopted for the first law of thermodynamics, positive work signifies that work is done on the water. 6. INTERPRET This problem involves finding the mechanical work that is required to raise the given mass of water by. C, which we can calculate by using the first law of thermodynamics. DEELOP The change in the internal energy of the water is Δ U mcδ T, where ΔT. C. K and the work done on it is W 9. kj. Inserting this into the first law of thermodynamics (Equation 8.) gives Q ΔU W mcδt W. For part (b), note that if the water had been in perfect thermal isolation, no heat would have been transferred, so Q and we can solve for the work W as for part (a). ELUTE (a) Inserting the given quantities into the expression for the heat transferred gives Q mcδt W.5 kg 4.84 kj/ kg K. K 9. kj.7 kj (b) If Q, then the work needed is W mcδ T.5 kg 4.84 kj/ kg K. K 6. kj SSESS The fact that Q < for part (a) indicates that heat is transferred from the water to the environment, as expected. 7. INTERPRET The system of interest is the gas that undergoes expansion. The problem involves calculating the change in internal energy of a system, so the first law of thermodynamics comes into play. DEELOP The heat added to the gas is Q Pt (4 W)(5 s) J. In addition, the system does 75 J of work on its surroundings, so the work done by the surroundings on the system is W 75 J. The change in internal energy can be found by using the first law of thermodynamics given in Equation 8.. ELUTE Using Equation 8. we find Δ U Q+ W J 75 J 5 J SSESS Since Δ U >, we conclude that the internal energy has increased. 8- Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

2 8- Chapter 8 8. INTERPRET This problem involves calculating the rate of heat flow into the system given the rate at which the internal energy is increasing and the rate at which the system is doing work on its surroundings. DEELOP The dynamic form of the first law of thermodynamics (Equation 8.) is dq du dw dt dt dt The problem states that du/dt 45 W and dw/dt 65 W (where the negative sign is appropriate because the system is doing work). ELUTE Inserting the given quantities gives dq dt 45 W + 65 W W SSESS The rate of heat flow into the system is positive because the system is doing work on its surroundings and increasing its internal energy. 9. INTERPRET This problem is about heat and mechanical energy, which are related by the first law of thermodynamics. The system of interest is the automobile engine. DEELOP Since we are dealing with rates, we make use of the dynamic form of the first law of thermodynamics, Equation 8.: du dq dw + dt dt dt If we assume that the engine system operates in a cycle, then du/ dt. The engine s mechanical power output dw/dt can then be calculated once the heat output is known. ELUTE The above conditions yield ( dq/ dt ) out 68 kw and ( dw dt ) ( dq dt ) in gives or dw dq dq dq dq dw dt dt dt dt dt.7 dt dw dt ( dq dt) out in out / out 68 kw 4 kw (.7).7 /.7 /. Equation 8. then SSESS We find the mechanical power output dw/ dt to be proportional to the heat output, ( dq/ dt ) out. In addition, dw/ dt also increases with the percentage of the total energy released in burning gasoline that ends up as mechanical work. The mechanical output is negative because the system is doing work on its surroundings. Section 8. Thermodynamic Processes. INTERPRET This problem involves an ideal gas that changes from an initial pressure-volume state to a final pressure-volume state by traversing the given pressure-volume curve. From this, we are to find the work done by the gas during this process. DEELOP The work done by the gas is given by the negative of Equation 8., which is W by gas p d This is the area under the line segment B in Figure 8.9. We can find this area from simple geometrical arguments, without having to invoke calculus. ELUTE The work done by the gas is the area of the trapezoid under line B, which is Wbygas W ( P + P )( ) ( P + P )( ) P Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

3 Heat, Work, and The First Law of Thermodynamics 8- SSESS so The work can also be obtained from Equation 8.. On the path B, p P+ ( )( P P) ( ) P P Wbygas p d P ( ) + ( ) P.. INTERPRET The expansion of the ideal gas involves two stages: an isochoric (constant-volume) process and an isobaric (constant-pressure) process. We are asked to find the total work done by the gas. DEELOP For an isochoric process, Δ so W (see Equation 8. and Table 8.). On the other hand, for an isobaric process, the work done is W pδ which is the negative of Equation 8.7 because we are interested in the work done by the gas (Equation 8.7 gives the work done on the gas). ELUTE Summing the two contributions to find the total work gives W W + W p p p tot SSESS In the p diagram of Fig. 8.9, the area under C is zero, and that under CB, a rectangle, is p. The work done by the gas is the area under the p curve.. INTERPRET This problem involves the isothermal (i.e., the temperature is constant) expansion of a balloon as it rises. We are asked to find the work done by the helium as the balloon rises. DEELOP During an isothermal expansion, the work done by a given amount of ideal gas is W nrt ln which is the negative of Equation 8. because that form expresses the work done on the gas. ELUTE Inserting the given quantities yields W (. mol) 8.4 J/ ( mol K) ( K) ln(5.). kj SSESS The result is positive because positive work is done by the gas to expand its volume.. INTERPRET The constant temperature of K indicates that the process is isothermal. We are to find the increase in volume of the balloon and the work done by the gas, given the pressure difference in experiences. DEELOP We assume the gas to be ideal and apply the ideal-gas law given in Equation 7.: p nrt. For an isothermal process, T constant, so we obtain p p, from which we can find the fractional volume increase. The total work done by the gas can be calculated using the negative of Equation 8.4: W nrtln because we are interested in the work done by the gas, not on the gas. ELUTE (a) For the isothermal expansion process, the volume increases by a factor of p kpa 4 p 75 kpa (b) Using Equation 8.4, the work done by the gas is 4 W nrtln (. mol) 8.4 J/ ( mol K) ( K) ln J to two significant figures. SSESS Because >, we find the work to be positive, W >. This makes sense because the gas inside the balloon must do positive work to expand outward. 4. INTERPRET This problem involves an ideal gas that we compress in an isothermal process to half its volume. We are to find the work needed to do this. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

4 8-4 Chapter 8 DEELOP In an isothermal compression of a fixed quantity of ideal gas, work is done on the gas so we use Equation 8.4, W nrtln( / ), where and T K for this problem. ELUTE Inserting the given quantities into Equation 8.4 gives W nrtln /.5 mol 8.4 J/ mol K K ln / + 4. kj. SSESS The work is positive because positive work must be expended by an external agent other than the gas to compress the gas. 5. INTERPRET The thermodynamic process here is adiabatic, so no heat flows between the system (the gas) and its environment. We are to find the volume change needed to double the temperature. DEELOP In an adiabatic process, Q, so the first law of thermodynamics (Equation 8.) becomes Δ U W. The temperature and volume are related by Equation 8.b: T constant T The temperature doubles, so T T and.4, so we can solve for the fractional change in volume. ELUTE From the equation above, we have /( ) T T T T Thus, for the temperature to double, the volume change must be /( ) /(.4 ) T.77 T SSESS We see that increasing the temperature along the adiabat is accompanied by a volume decrease. In addition, since p constant, the final pressure is also increased: p p p p 6. INTERPRET The thermodynamic process here is adiabatic, so no heat flows between the system (the gas) and its environment. We are to find the temperature increase as the gas volume is reduced. DEELOP In an adiabatic process, Q, so the first law of thermodynamics (Equation 8.) becomes Δ U W. The temperature and volume are related by Equation 8.b: T T The final volume is /4 of the initial, so /4 and.4, so we can solve for the change in temperature. ELUTE From the equation above, we have.4 T T T T (9K)(4) 57 K Thus, the temperature change is Δ T T T 57 K 9K 6 K SSESS We see that decreasing the volume along the adiabat is accompanied by a temperature increase. In addition, since p constant, the final pressure is also increased: p p p ( 4) p 7. INTERPRET This problem involves an ideal gas that we compress in an isothermal process to half its volume. We are to find the work needed to do this. DEELOP In an isothermal compression of a fixed quantity of ideal gas, work is done on the gas so we use Equation 8.4, W nrtln( / ), where /.5 for this problem. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

5 Heat, Work, and The First Law of Thermodynamics 8-5 ELUTE Inserting the given quantities into Equation 8.4 gives W nrt p 5 6 ln / ln / (. Pa) 6.8 m ln.5. J. SSESS The work is positive because positive work must be expended by an external agent other than the gas to compress the gas. Section 8. Specific Heats of an Ideal Gas 8. INTERPRET This problem involves finding the molar specific heat at constant volume and constant pressure for the given gas mixture, which contains the given quantities of diatomic O and monatomic r. DEELOP This problem in similar to Example 8.5, so we will use the same approach as outlined there. ELUTE Because O has 5 degrees of freedom, so its average energy per molecule is 5kT/. For r, the average energy per molecule is kt/ because it is monatomic. The total internal energy is therefore 5 5 U no RT n r RT +.5 mol. mol RT.75 mol RT + From Equation 8.6, the molar specific heat at constant volume is (.75 mol) ( + ) du R C.R ndt.5. mol to two significant figures. From Equation 8.9, the molar specific heat at constant pressure is CP C + R.R SSESS We find that the molar specific heat at constant volume is between that of a diatomic molecule (.5 R) and a monatomic molecule (.5 R), as expected since we have a mixture of these two types of molecules. 9. INTERPRET The problem is about the specific heat of a mixture of gases. We want to know what fraction of the molecules is monatomic, given the its specific-heat ratio. DEELOP The internal energy of a mixture of two ideal gases is U fne+ fne where f i is the fraction of the total number of molecules, N, of type i, and E i is the average energy of a molecule of type i. Classically, E g( kt ), where g is the number of degrees of freedom (the equipartition theorem). From Equation 8.6, the molar specific heat at constant volume is du d C nr fg T fg T R( fg fg) ndt dt + + Suppose that the temperature range is such that g for the monatomic gas, and g 5 for the diatomic gas, as discussed in Section 8.. Then C R( f+ 5f) R(.5 f) where f f since the sum of the fractions of the mixture is unity. Now, C can also be specified by the ratio CP C + R R + C C C R C Equating the two expressions allows us to solve for f. ELUTE Solving, we find.5 f f 57.7%.5 SSESS From the equation above, we see that the specific-heat ratio can be written as +.5 f Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

6 8-6 Chapter 8 In the limit where all the gas molecules are monatomic, f, and.67. On the other hand, if all the molecules are diatomic, then f and the specific-heat ratio is.4. The equation yields the expected results in both limits.. INTERPRET For this problem, we are to find the specific-heat ratio of a gas consisting of the given ratio of different gas molecules. DEELOP By generalizing the result of the previous problem, the molar specific heat of a mixture of three gases is C fc + f C + f C. For each gas and the mixture, use C R ( ) to obtain ( ) f ( ) + f ( ) + f ( ) which we can evaluate to find. ELUTE When the data specified in the problem is inserted, one gets SSESS Because we have some triatomic molecules, the specific-heat ratio is less than that for a pure diatomic gas (see previous problem).. INTERPRET The thermodynamic process is adiabatic, and we want to know the temperature change when work is done on a monatomic gas and a diatomic gas. DEELOP In an adiabatic process, Q, so from the first law of thermodynamics (Equation 8.) Δ U W, where W is the work done on the gas. From Equation 8.6, Δ U nc Δ T, the change in temperature is PROBLEMS ΔU W Δ T nc nc If the work done per mole on the gas is W n.5 kj/mol, then Δ T (.5 kj/mol) C ELUTE (a) For an ideal monatomic gas, C R 8.4 J/ ( mol K ), so Δ T K. 5 (b) For an ideal diatomic gas (with five degrees of freedom), C R so Δ T K. SSESS Since the diatomic gas has a greater specific heat C, its temperature change is less than that of the monatomic gas.. INTERPRET The constant temperature of 44 K indicates that the process is isothermal. We are given the amount of work done by the gas, and are asked to find the heat it absorbs and how many moles of gas there are. DEELOP pply the ideal-gas law given in Equation 7.: p nrt. For an isothermal process, T constant, so we obtain p p. Since U Δ for an isothermal process, the heat absorbed is the negative of the total work done on the gas (Equation 8.4). Q W nrtln For this problem, the gas does. kj of work on its surroundings, so the surroundings do. kj of work on the gas, so W. kj. ELUTE (a) Using the equation above, the heat absorbed is Q W. kj. (b) Solving the expression above for the number n of moles gives W. kj n.9 mol RT ln ( / ) 8.4 J/ ( mol K) ( 44 K) ln SSESS The heat absorbed by the gas is equal to the work done by the gas on its surrounding as it expands, and there is no change in temperature. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

7 Heat, Work, and The First Law of Thermodynamics 8-7. INTERPRET We re asked to find the rate that heat is produced in the body when cycling. This involves the first law of thermodynamics. DEELOP We re dealing with the rates of work and heat production, so we ll use Equation 8.: du dq + dw. dt dt dt If the body is releasing stored food energy, that corresponds to a decrease in the body s internal energy: du / dt 5 W. Likewise, the mechanical power quoted is for work done by the body, so dw / dt W. We are looking for the rate at which the body produces heat, which is technically heat that it is losing ( dq). ELUTE The rate of heat production is the negative heat absorbed, so dq du dw + ( 5 W) + ( W) 8 W dt dt dt SSESS ll the signs can be confusing, but essentially the body burns stored energy and some of it is used to do work and the rest is released as heat. 4. INTERPRET This problem deals with isothermal compression, so the temperature is constant in this process. We are to find the work done to compress the given quantity of ideal gas from.5 L to. L. DEELOP pply Equation 8.4, Q W nrtln where W is the work done by the gas,.5 L, and. L. Since we do 6 J of work on the gas to compress it, W 6 J, so we can solve this expression for the temperature T. ELUTE The temperature of the ideal gas is W 6 J T 9 K nrln( ) (.5 mol) 8.4 J/ ( mol K) ln(./.5) SSESS This gas is about K below room temperature. 5. INTERPRET ssume the air inside the spherical bubble behaves like an ideal gas at constant temperature, so the process is isothermal. We need to find the diameter of the bubble at maximum pressure and the work done on the gas in compressing it. DEELOP pply the ideal-gas law given in Equation 7.: p nrt. For an isothermal process, T constant, which leads to p. p Since the volume of a spherical bubble of diameter d is 4π d πd 6 the relationship between the diameter and the pressure is πd πd p p 6 6 / d p d p ELUTE (a) Using the equation above, we find the diameter at the maximum pressure to be / / p ( ) mm of Hg d d (.5 mm ).49 mm p ( ) mm of Hg (b) The work done on the air is given by Equation 8.4, or p p W nrtln p ln p ln SSESS on air p p. kpa 885 mm of Hg π ( 84 mm of Hg) (.5 mm) ln 76 mm of Hg 6 84 mm of Hg.7 μj Positive work is done by the blood in compressing the air bubble. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

8 8-8 Chapter 8 6. INTERPRET This problem involves compressing a gas in an isothermal process, so the temperature is constant. We are to find how much work it takes to reduce the volume by a factor of given the work it takes to decrease the volume by a factor of. DEELOP pply Equation 8.4 to both situations. To compress a gas by a factor of two, we have W nrtln and to compress it by a factor of gives W nrtln Given that T is constant, we can take the ratio of these expressions to find the work W needed to compress the gas by a factor of. ELUTE The work needed to compress the gas by a factor of is W ln ln W (.5 kj) 6.7 kj ln ln SSESS Note that the work required is logarithmic in volume. 7. INTERPRET The thermodynamic process here is adiabatic, with no heat flowing between the system (the gas) and its environment. We are to find the specific-heat ratio given the fraction increase in pressure of the gas. DEELOP In an adiabatic process, Q and the first law of thermodynamics becomes Δ U W. The pressure and volume are related by Equation 8.a: P constant. This implies p p p p so we can solve for. ELUTE Taking the natural logarithm on both sides of the above to solve for, we obtain p ln( p p) ln(.55) ln ln.5 p ln( ) ln SSESS The value of indicates that gas consists of polyatomic molecules (see Problems 9 and ). 8. INTERPRET The problem involves compressing a gas adiabatically. DEELOP For an adiabatic process, the initial pressure and volume are related to the final pressure and volume through Equation 8.a: p. We can use this to rewrite Equation 8. for the work done on the gas in going from the initial to the final volume: p p p W ELUTE (a) The final pressure in this case is 6.5 L ( 98.5 kpa) 7 kpa 4.8 L (b) The work done on the gas in compressing it from 6.5 L to 4.8 L is p p kpa 6.5 L 6.5 L W 69 J L SSESS The volume decreases by a factor of.5, but the pressure increases by a factor of.76. The pressure increases due to both the increase in the density (N/) and an increase in the internal energy of the gas ( Δ U nc Δ T). 9. INTERPRET This problem involves a cyclic process. The three processes that make up the cycle are: isothermal (B), isochoric (BC), and isobaric (C). We are given the pressure at point and are to find the pressure at point B and the net work done on the gas..4 Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

9 Heat, Work, and The First Law of Thermodynamics 8-9 DEELOP long the isotherm B T constant, so the ideal-gas law (Equation 7., p nrt) gives p. pbb For an isothermal process, the work W done on the gas is (Equation 8.4): W nrtln ELUTE (a) If B is an isotherm, with 5 L and L, then the ideal-gas law gives 5 PB P ( 6 kpa) kpa B (b) The work W done on the gas in the isothermal process B is B. L B WB nrt ln p ln ( J) ln 48 J 5. L The process BC is isochoric (constant volume) so W BC. The process C is isobaric (constant pressure) so the work done by the gas is (see Table 8.) W p 6 kpa 5. L. L 4 J C C or the work done on the gas is W C 4 J. The total work done by the gas is the sum of these three contributions, or WBC WB + WBC + WC 48 J + 4 J 4 J 4 J to two significant figures. SSESS Since the process is cyclic, the system returns to its original state, there s no net change in internal energy, so Δ U. This implies that Q W BC 4 J. That is, 4 J of heat must come out of the system. 4. INTERPRET This problem is similar to the previous one, except that the isotherm curve B is now to be considered as an adiabat. We are given the specific-heat ratio and are to find the pressure change in going from point to point B, and the net work done by the cycle. DEELOP For an adiabat, Equation 8.a is valid. pplying this to points and B, and taking the ratio, gives p B pb which allows us to find the pressure at point B. The calculation for part (b) then proceeds as for Problem 8.9, with the help of Equation 8., which gives the work done on the gas in an adiabatic process. ELUTE (a) The pressure at point B is 5. L pb p ( 6 kpa) 57 kpa B. L to two significant figures. (b) The work done on the gas between points and B is W B B ( 57 kpa)(. L) ( 6 kpa)( 5. L) P B B P 678 J.4 The process BC is isochoric (constant volume), so W BC. The process C is the same as for Problem 8.7, so WC 4 J. Summing all three contributions gives a net work done on the gas of WBC WB + WBC + WC 678 J + 4 J 44 J. to two significant figures. SSESS Because the net work done on the gas is negative, the gas does work on its environment. 4. INTERPRET We identify the thermodynamic process here as adiabatic compression. DEELOP In an adiabatic process, Q, and the first law of thermodynamics becomes Δ U W. The temperature and volume are related by Equation 8.b: T constant.4 Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

10 8- Chapter 8 From the equation above, we obtain T T T T where / is the compression ratio (for T and at maximum compression). ELUTE Inserting the values given gives.4 T T ( K )( 8.5 ) 7 K 44 C SSESS Note that the temperature T appearing in the gas laws is the absolute temperature. The higher the compression ratio /, the greater the temperature at the maximum compression, and hence a higher thermal efficiency. 4. INTERPRET This problem involves an adiabatic process that results in a change of pressure and volume. We are to find the fractional change in volume given that the pressure increases by a factor of. DEELOP pply Equation 8.a, p constant. Using subscripts and to indicate before and after the process, respectively, this gives p constant p p p which we can solve for the fractional change in volume, using.4 and p p. ELUTE Inserting the given quantities into the expression above gives p.4 (.5).6 p SSESS Thus, the new volume is almost 5% of the original volume, which seems reasonable given that the pressure has doubled. 4. INTERPRET We identify the thermodynamic process here as adiabatic compression. We are to find the temperature and pressure in the cylinder when it is at maximum compression. DEELOP In an adiabatic process, Q, and the first law of thermodynamics becomes Δ U W. Because we are dealing with an adiabatic process, the temperature and volume are related by Equation 8.b: T constant pplying this expression before (subscript ) and after (subscript ) compression gives T T T T where. is the compression ratio (for T and at maximum compression) and.4. In addition, since p constant (Equation 8.a), the final pressure is p p( ). ELUTE (a) Substituting the values given in the problem statement, we find the air temperature at the maximum compression to be (b) The corresponding pressure is T T.4.4 ( K)(.) 8 K p p. kpa..6 MPa 5.8 atm SSESS The higher the compression ratio /, the greater the temperature and pressure at the maximum compression, and hence, a higher thermal (fuel) efficiency. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

11 Heat, Work, and The First Law of Thermodynamics INTERPRET The thermodynamic process here is adiabatic, so no heat flows between the system (the gas) and its environment. We are to find the change in temperature and volume as the pressure is reduced. DEELOP In an adiabatic process, Q, so the first law of thermodynamics (Equation 8.) becomes Δ U W. The temperature, pressure and volume are related by Equation 8.: T T, p p The initial pressure is p. atm, and the final pressure is p.4 atm. With 5/, we can solve for the new volume and temperature. ELUTE (a) From the equation above, we have /5 /. atm ( p p) (.75 m ).4 m.4 atm (b) The final temperature is (5/).4 m.75 m T T (85 K) 85 K 88 C SSESS We see that decreasing the pressure along the adiabat is accompanied by a temperature decrease. In addition, since p constant, the final volume increases. 45. INTERPRET This problem involves an adiabatic compression of a gas, for which we know the initial and final temperatures. We are to find the ratio of the final pressure to the initial pressure of the gas. DEELOP For an adiabatic process, the temperature and pressure are related by Equation 8.a: p constant Rearranging this equation and using the ideal-gas law (Equation 7., p nrt) gives ELUTE constant /( ) /( ) p p p p nrt pt p T Inserting the given quantities into the expression above for temperature gives p p 5/ (5/) T 9 K 54 T 8 K SSESS The pressure increase is enormous in order to accompany the increase in temperature along the adiabat. The corresponding change in volume change is /5 p.96 p 54 The final volume is only about % of the original. 46. INTERPRET This problem explores how different ways of compressing volume (isothermal, adiabatic, or 5 isobaric) affects the internal energy of the system. The internal energy of n moles of diatomic gas is U nrt, by equipartition theorem. DEELOP In an isothermal process, the temperature T is kept constant, so Δ U. In an isobaric process, p constant, and T / T /. Finally, in an adiabatic process, Q and T T ELUTE (a) For isothermal compression, ΔT, so Δ U. There is no change in internal energy. (b) For isobaric compression, U 5 / nrt T. U 5 nrt/ T (c) For adiabatic compression, 7/5 for diatomic gas, U 5 nrt / T.4. U 5 nrt/ T Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

12 8- Chapter 8 SSESS s the gas volume is halved, the internal energy of the gas decreases if compression is isobaric, but increases if the compression is adiabatic. There is no change if the compression is isothermal. 47. INTERPRET In the problem we compare the work done in adiabatic compression to the work done in isothermal compression. DEELOP For an adiabatic process, the initial pressure and volume are related to the final pressure and volume through Equation 8.a: p. We can use this to rewrite Equation 8. for the work done on the gas in going from the initial to the final volume: p p p Wadiabatic On the other hand, for an isothermal process, the work W done on the gas is (Equation 8.4): Wisothermal nrtln p ln p ln ELUTE (a) For / /, and 5/, from the above expressions, we find the ratio of the work done for the two processes to be Wadiabatic /.7 Wisothermal ( )ln( / ) (/)ln() (b) The extra work goes into internal energy, raising the temperature of the gas. SSESS For diatomic gas where 5/, the ratio would be Wadiabatic.4.5 Wisothermal ( )ln( / ) (.4)ln() 48. INTERPRET This problem involves first an isothermal expansion, then an isobaric compression, then finally an isochoric process to return to the original state. We are to draw a p diagram of this process, then calculate the work done over the entire cycle, then finally, find the heat transfer between the given stages. DEELOP The isothermal expansion is along the curve B in the figure below, which is given by P constant. Because the gas expands, it starts at the lower volume at point, and expands to occupy the greater volume B at point B. For the isobaric compression, the gas follows the horizontal line BC, with the pressure remaining constant at P B P C. Finally, for the isochoric process, the gas follows the vertical line C, with the volume remaining constant at C. To find the total work done on or by the gas during this cycle, sum the work done over each interval, Wtot WB + WBC + WC where we consider work done on the system to be positive. For the isotherm (process B), ΔU and the heat transferred to the system is Q B 5 J. From the first law of thermodynamics (Equation 8.), this gives W B Q B 5 J. For process BC, the work done on the gas is W BC J, and W C because the volume of the gas does not change (see Equation 8.7). To find the heat that enters or leaves the system during the stages BC, recall that the change in internal energy is zero for the whole cycle, so Δ Utot Qtot + Wtot and that the total heat is composed of two parts: Qtot QB + QBC so we can solve for Q BC. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

13 Heat, Work, and The First Law of Thermodynamics 8- ELUTE (a) See figure above. (b) Summing the various contributions to the work gives Wtot WB + WBC + WC 5 J+ J+ J J so the gas does J of work on its environment. (c) The heat transferred to the system is Qtot QB + QBC Wtot QBC Wtot QB J 5 J J so J of heat is transferred out of the gas during the process BC. SSESS The total heat transferred to the system over the cycle is Q tot 5 J J J, which provides the energy for the J of work done by the system over this same cycle. 49. INTERPRET This problem explores how different ways of adding heat (isothermal, isochoric, or isobaric) affects the final temperature of the system. Starting with the given quantity of gas at the given temperature, we are to find the work done by the gas upon adding heat to the gas via these different processes. DEELOP In an isothermal process, the temperature T is kept constant, so Δ U. The first law of thermodynamics (Equation 8.) therefore gives Q W. In an isochoric process, Δ and W (see Equation 8.7), so the first law of thermodynamics gives QΔ U nc Δ T. Finally, in an isobaric process, Δp and Q ncpδ T n( C + R) Δ T Use these expressions to solve for ΔT and W for each case. ELUTE (a) With ΔU, W Q.75 kj. For an isothermal process, the temperature is constant so T 55 K. (b) Solving the expression above for ΔT, we find for an isochoric process Δ Q.75 kj T 4 K nc (.5 mol)( 5R ) Therefore, T 55 K +Δ T 79 K. Because the volume does not change, W. (c) In an isobaric process, Q Q.75 kj Δ T 7 K ncp n( C + R) (.5 mol)( 5R + R) and T 55 K + 7 K 7 K. The work done by the gas is R R.75kJ W p Δ nr Δ T Q Q 5 J C p ( 7R ) 7 SSESS Comparing all three cases, we find ΔT : isothermal < isobaric < isochoric W: isochoric < isobaric < isothermal The results agree with that illustrated in Table INTERPRET This problem is an exercise to show the functional relationship in terms of pressure and volume between an adiabatic process and an isothermal process. DEELOP The equations of an adiabat and an isotherm, passing through the point (, p ) in the p diagram are and p p p p respectively. Differentiate these expressions and evaluate the results at (, p ) to find the desired relationship. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

14 8-4 Chapter 8 ELUTE Differentiating the above expression for an adiabat and evaluating the result at and p p gives dp p d adiabat Differentiating the expression above for an isotherm and evaluating the result at gives dp p p d isotherm Comparing the two results gives dp dp d d p ( ) p p adiabat SSESS We have found the expression given in the problem statement. Because >, we see that the slope of the p curve for an adiabat is greater than that for an isotherm, so the pressure of an adiabatic process changes more rapidly with volume compared with an isothermal process, as shown in Fig INTERPRET The problem involves a cyclic process with three separate stages: adiabatic, isochoric, and isothermal. We are given the initial volume and pressure of the gas, and its adiabatic exponent, and are to find the pressure at points B and C and the net work done on the gas. DEELOP For the adiabatic process B, Q and the first law of thermodynamics becomes Δ U W. The pressure and volume are related by Equation 8.a: p constant, which gives p pbb pb p B Point C lies on an isotherm (constant temperature) with, so the ideal-gas law (Equation 7.) yields p pc C To find the net work done on the gas, sum the contributions from each stage of the cycle. The contribution W B may be found from Equation 8. for an adiabatic process: pbb p WB For an isochoric (constant volume) process W BC (see Equation 8.7), and for the isothermal process WC nrt ln C ELUTE (a) From the equation above, the pressure at point B is (b) The pressure at point C is p isotherm.67 B p B 5 kpa 4 kpa 5 kpa pc p 8 kpa. C (c) The net work done by the gas is W W + W + W. BC B BC C The work W B for the adiabatic segment is p ( ) ( ) B B p 9.9 kpa. m 5 kpa. m WB 94 kj.67 WBC is for an isochoric process and equals zero. Finally, WC is for an isothermal process and is WC nrt ln ( 5 kj) ln 75 kj C Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

15 Heat, Work, and The First Law of Thermodynamics 8-5 Summing these contributions gives WBC WB + WBC + WC 94 kj kj 8 kj so the work done on the gas 8 kj. SSESS Since the process is cyclic, the system returns to its original state, there s no net change in internal energy, so ΔU. This implies that Q WBC 8 kj. That is, 8 kj of heat must come out of the system. 5. INTERPRET This problem involves a cyclic process with an adiabatic compression, an isobaric compression, and an isothermal expansion. We are to find the net work done on the gas and its minimum volume. DEELOP The p diagram for this problem is different than that of Figure 8.4; only point is the same (see figure below). Note that p 4 J and B. To find the net work done on the gas, sum up the contributions from each stage. For the adiabatic stage B, the work done on the gas is given by Equation 8.: pbb p WB For the isobaric process BC, the work done on the gas is given by Equation 8.7: W p Δ p BC B B C B ia the adiabatic process, we can find the product p B B (using Equation 8.), which gives pb p B B B B p p ia the isothermal process C, we know that C is at the same temperature as. We also know that p C p B because BC is isobaric. Thus, from the ideal-gas law (Equation 7.), we have pcc pbc p Therefore, the work W BC is W p B BC The work done on the gas for the isothermal process C is given by Equation 8.4, which gives (using the idealgas law p nrt) WC p ln C The minimum volume of the gas occurs at point C, and is given by C B B pb p Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

16 8-6 Chapter 8 ELUTE (a) Summing the work done on the gas for each process gives W ( ) ( 4 J)(.4 ) p p.4 B B B B p 9.5 J where we have used the expression above relating p B B to p from the adiabatic process. Continuing, we have for process BC.4 WBC p ( 4 J)( ) 7.8 J B and for process C we have W p p C ln ln 4 J.4 ln 88. J C B Summing these, the total work done on the gas is W tot 9.5 J+ 7.8 J 88. J 59.J, or about 59 J. (b) The minimum volume is SSESS.4 B C B (. L).5 L The total work is positive, so work is done on the gas in this cycle. 5. INTERPRET We are to find the work done in the given heat cycle, which consists of an isochoric doubling in pressure, and adiabatic compression, an isochoric cooling to K, and an isothermal expansion. DEELOP The net (or total) work done on the gas is the sum of the work done for each process in the cycle (see figure below), which we give here: (B) It is heated at constant volume until the pressure is doubled. Because Δ, W B (see Equation 8.7). (BC) It is compressed adiabatically until its volume is the initial value. This is analogous to the process B of 4 Problem 8.46, so the result is ( ) C B WBC pbb where B / C /4 and p B B 8 J (see figure below). (CD) It is cooled at constant volume to a temperature of K. No work is done because Δ, so W CD. (D) It is expanded isothermally until it returns to the original state. This is analogous to the process C of Problem 8.46, so we use that result: WD p ln D where / D 4 and p 4 J (see figure). Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

17 ELUTE Summing the nonzero contributions to the work gives Heat, Work, and The First Law of Thermodynamics 8-7 B C tot BC D B B ln D.4 W W + W p p 4 ( 8 J) ( 4 J) ln( 4) 9 J.4 SSESS Work is done on the gas. From the figure above, we see that the area under the entire curve is negative, because the gas goes around the cycle counterclockwise. s per Figure 8.6, the work done by the gas is the negative of the area under the p curve. 54. INTERPRET This problem is similar to the preceding two problems. It involves again a thermodynamic cycle, this time involving an isothermal compression, followed by an isochoric increase in pressure, then an adiabatic expansion. We are to find the total work done on or by the gas over this cycle. DEELOP We begin by drawing the p curve for this cycle (see figure below). Sum the contributions to the work from each process in the cycle. For the isothermal process B, apply Equation 8.4 and the ideal-gas law (Equation 7., p nrt). The work done on the gas is B WB p ln where p.5 kj and B / / (see figure below). For the isochoric process BC, no work is done because Δ (see Equation 8.7). For the adiabatic process C, the work done is where p.5 kj and / C /. W p C ( ) C ELUTE Summing the nonzero contributions to the work gives ( ) C B Wtot WB + WC p ln + p.67.. ( 5 J) ln( ) + ( 5 J) 656 kj.67 SSESS The work is negative, which means that the gas does work on the environment, which is consistent with the clockwise direction of the cycle about the p curve. 55. INTERPRET We re asked to derive the relation between pressure and temperature in an adiabatic process. DEELOP We just need to combine Equations 8.a and 8.b: p constant, and T constant. ELUTE Isolating the volume in both equations and matching the exponents gives ( ) p T p T constant ( ) constant ( ) p ( ) constant T Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

18 8-8 Chapter 8 The constant term is just a place-holder. You can raise the constant to a power and it s still just a constant. Or you can multiply the constant by another constant, and the result is still just a constant. SSESS For >, the result says the pressure and temperature will increase or decrease together during an adiabatic process. 56. INTERPRET The problem involves a cyclic process with three separate stages, which are (in order) adiabatic, isochoric, and isothermal. We are to find the total work done in the process. s seen in Problems , there will be net work on the gas by the environment because the system proceeds in a counterclockwise manner around the cycle. DEELOP We begin by sketching the p curve for this cycle (see figure below). The total, or net, work done is the sum of the work done in the individual processes of the cycle. For the adiabatic process B, Q, and the first law of thermodynamics (Equation 8.) becomes Δ U W. The pressure and volume are related by Equation 8.a: p constant, and the work done by the gas is given by Equation 8.: pbb p WB Since Δ BC for the isochoric process, W BC. Similarly, for an isothermal process, the work done is (Equation 8.4) W nrt ln p ln C C C where we have used the ideal-gas law (Equation 7.) p nrt. The minimum volume attained is B C, which is related to by the adiabatic compression, because p constant for an adiabatic process (Equation 8.a), we have / p C B pb where p /p B / (see figure below) and 5 L. ELUTE The solutions below are presented in the reverse order. (a) The work done on the gas is the sum of the work done on the gas during each process: p B B p Won gas WBC WB + WBC + WC p ln C ( p pb) ln( pb p) p.67 ln ( ) 5 kpa 5 L J to two significant figures, and where we have used the relationship between, B, and C derived for part (b) to find the minimum volume. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

19 Heat, Work, and The First Law of Thermodynamics 8-9 (b) Inserting the volume and pressure in to the expression for the minimum volume gives / /.67 p C B ( 5 L) L p B (c) See the figure above for the p cycle. SSESS Since the process is cyclic, the system returns to its original state and there is no net change in internal energy, so Δ U. This implies that Q WBC J. That is, J of heat must come out of the system. 57. INTERPRET The pump handle is pressed rapidly, so you can assume that there s no time for heat to flow into or out of the gas in the pump. This means that process is adiabatic. DEELOP You want to check that the temperature rise in the pump is less than 75 C. Equation 8.b tells you how to relate the final temperate and volume to the initial temperature and volume: T T. Therefore, the temperature rise is Δ T T T T o The gas in the cylinder is air (.4) initially at T 8 C 9 K. ELUTE Since the cross-sectional area of the cylinder remains unchanged during the compression, the volume ratio is just equal to the ratio in the cylinder s height:.4 cm Δ T T 9 K 9 K 6 cm This does not meet your 75 C criteria. SSESS There s no obvious way to avoid this temperature rise. The cylinder can be made of a material with a high thermal conductivity, such as aluminum, so that heat can flow out as fast as possible. 58. INTERPRET We re given experimental data and are asked to estimate the work involved in inflating a human lung. DEELOP We want calculate the work done by the gas in expanding the lungs: Δ W pδ. In the given graph, we can take the volume change between successive points as Δ, and the pressure at the midpoint between successive points as the average pressure p. To find the total work done, we sum over these small individual expansions. ELUTE There are seven main volume expansions that correspond to pressure increases of 5 kpa. We add the work involved in each expansion to get the total work. (Note: to save space, we leave off the units until the end of the sum): W pδ ( p i + pi )( i i ) (.5 + 5)( 5 ) + ( 5 + )( 6 5) + ( + 5)( 5 6) + ( 5 + )( 8 5) + ( + 5)( 8) + ( 5 + )( 5 ) + ( + 5)( 55 5 ) kpa ml J SSESS It may seem counterintuitive that the pressure increases as the volume increases, but in this case air is rushing into the lungs from the outside. When a person breathes, their diaphragm contracts to open up the lung cavity. This lowers the pressure on the lungs, so they inflate like two balloons. The air rushing into the lungs does the work to expand the lung volume; the diaphragm is only indirectly responsible. fterwards, the diaphragm expands to squeeze on the lungs, which increases the pressure and thus forces air out. 59. INTERPRET For this problem, we are given the heat transferred from a monatomic gas to the surroundings and the change in internal energy of the gas, and we are to find the work done in the process. DEELOP pply the first law of thermodynamics, Equation 8., ΔU Q + W, where Q 5 kj. The internal energy change can be calculated using the constant-volume specific heat for a monatomic gas, which is C R/ (see Equation 8.). Thus, Δ U nc Δ T where n mol and ΔT 6 K. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

20 8- Chapter 8 ELUTE Combining the above expressions and inserting the given quantities gives W Δ U Q nc Δ T Q ( mol) 8.4 J/ ( mol K) ( 6 K) + 5 kj 57 kj SSESS The work is positive so the work is done on the gas. 6. INTERPRET The thermodynamic process here involves two stages: isothermal compression followed by an adiabatic compression. We are to find the final temperature of the gas. DEELOP During the first stage, the gas is compressed isothermally so ΔT and there is no change in the temperature of the system (i.e., T T ), but the volume is reduced from to /. During the next stage of adiabatic compression, the temperature and volume are related by Equation 8.b: T constant which we apply to the points and (i.e., before and after the adiabatic compression, respectively) to find T T T T where the final temperature is T. ELUTE Substituting the values given, we find T to be 7/5.4 / 5 /5 T T T 7 K 5 K SSESS Since T constant for an adiabatic process, the compression results in an increase of the final temperature. 6. INTERPRET This problem involves an adiabatic compression of a gas, for which we know the initial temperature and pressure and the final pressure. We are to find the final temperature of the gas. DEELOP For an adiabatic process, the temperature and pressure are related by Equation 8.a: p constant Rearranging this equation and using the ideal-gas law (Equation 7., p nrt) gives pt ( ) p p p p nrt constant another constant (/ ) pplying this to the gas both before and after the compression gives T T( P/ P) ELUTE Inserting the given quantities into the expression above for temperature gives (.) (/ ) kpa ( / ) ( 7K) 4K T T P P 4 kpa or. K to two significant figures. SSESS Because the gas is compressed adiabatically, its temperature rises. 6. INTERPRET The problem involves a cyclic process with three separate stages of the cycle: isochoric (B), isobaric (BC), and isothermal (C). We are to calculate the net work done on the gas over the entire cycle, and the heat transfer over segment B. DEELOP The work done on the gas in each segment of the cycle is summarized in Table 8.. For the isochoric process (path B), Δ so W. For the isobaric process (path BC), the work done is WBC pb( C B) Finally, for the isothermal process (C), the work done on the gas is given by Equation 8.4: W C nrtln C Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may

21 Heat, Work, and The First Law of Thermodynamics 8- Segment B is isochoric, so Q B nc ΔT (see Table 8.). To eliminate C, note that CP C R R R + C C C + C which follows from the definition of the adiabatic ratio (see discussion for Equation 8.a). Inserting this into the expression for the heat transfer gives nr( TB T) pbb p QB where the second equality follows from the ideal-gas law (Equation 7.), p nrt. ELUTE (a) Using the equations above, we obtain WB (isochoric) WBC pb( C B) ( 5 kpa)(. L 5. L). kj (isobaric) WC nrt ln p ln ( 5. kpa)( 5. L) ln( 5.) 4 J (isothermal) C C dding up all the contributions, the net work done on the gas is (b) The heat transferred is Q WBC WB + WBC + WC + J 4 J 598 J B ( 5 kpa 5 kpa )( 5. L ) p B B p.5 kj.4. Since Q B >, heat is transferred into the gas. SSESS t constant volume, the gas must be heated in order to raise its pressure. CP C + R R R + C C C C 6. INTERPRET This problem is similar to the preceding one, except that the order of segments in the cycle are changed. The cycle proceeds along the 5-K isotherm from 5 kpa to 5 kpa (C), then decreases its pressure to 5 kpa via an isochoric process (CD), then returns to the starting point via a 5-kPa isobar (D). We are to find the total work done on the gas for the complete cycle and the heat transferred in segment CD. DEELOP The formulas are the same as for the previous problem, but the initial and final point for each segment are different. The work done on the gas over segment C will be the opposite of the result we found for the segment C in Problem 8.56, so W C 4 J. The work done over segment CD is zero because the volume is constant (see Equation 8.7). The work done over segment D is W p ( ) BC D D The heat transfer can be calculated as per Problem 8.56, which gives pdd pcc QCD ELUTE (a) The total work is W W + W + W W p ( ) 4 J 5. kpa 5. L. L J CD C CD D C D D (b) The heat transfer over segment CD is ( 5 kpa 5 kpa)( L) p D D p C C QCD 5 J.4 Because Q CD <, we interpret the result as 5 J of heat transferred out of the gas. SSESS t constant volume, the gas must be cooled to lower its pressure. Copyright 6 Pearson Education, Inc. ll rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may