University Physics AI No. 10 The First Law of Thermodynamics Class Number Name I．Choose the Correct Answer 1. Which of the following processes must violate the first law of thermodynamics? (There may be more than one answer!) （ A,B,D ） (A) W > 0, Q < 0, and ?E int = 0 (B) W > 0, Q < 0, and ?E int > 0 (C) W > 0, Q < 0, and ?E int < 0 (D) W < 0, Q > 0, and ?E int < 0 (E) W > 0, Q > 0, and ?E int < 0 Solution: Applying the first law of thermodynamics WEQ +?= . 2. In which of the paths between initial state i and final state f in Fig.1 is the work done on the gas the greatest? ( D ) Solution: The work done by the gas is the area under the path, and the work done on the gas is the negative of the work done by the gas. 3. Consider the following processes that can be done on an ideal gas: constant volume, ?V =0; constant pressure, ?P =0; and constant temperature, ?T =0. (a) For which process does W=0? ( A ) (b) For which process does Q=0? ( D ) (c) For which process does W+Q=0? ( D ) (d) For which process does ?E int = Q? ( A ) (e) For which of these processes does ?E int = W? ( D ） (A) ?V =0 (B) ?P =0 (C) ?T =0 (D) Non of these Solution: Applying the first law of thermodynamics. 4. Which of the following processes is forbidden by the first law of thermodynamics? (There may be more than one correct answer!) （C） (A) An ice cube is placed in hot coffee; the ice gets colder and the coffee gets hotter. (B) Solid wax is placed in hot metal pan; the wax melts and the metal pan cools. (C) Cold water is placed in a cold glass; the glass gets colder and the water gets colder. (D) A student builds an automobile engine that converts into work the heat energy released when water changed to ice. (E) Dry ice can be made by allowing carbon dioxide gas to expand in a bag. Solution: Only （C）violates the first law of thermodynamics.. i f V P A B C D Fig.1 II. Filling the Blanks 1. A 0.500 kg block of ice at -10.0 °C is placed in a punch bowl with 4.00 kg of water (punch) at 20.0 °C. Assume no heat transfer to the surroundings and neglect the thermal effects of the punch bowl itself. Does all the ice melt? yes . If not, the quantity which is left is . The final temperature of the punch plus ice system is 281.53 K . Solution: To warm the ice from 10°C to 0°C takes a heat transfer to the ice system )J(10025.1)100(20505.0 4 1 ×=+××=?= TCmQ iceice If all the ice melts, it need heat transfer )J(106675.110335.35.0 55 2 ×=××== fice LmQ Let the water (punch) changes from 20°C to 0°C, the heat transfer is )J(103488.3)200(41864 5 3 ×?=?××=?= TCmQ waterwater So 321 QQQ <+ we can know all the ice will melt. As 0 321 <++ QQQ So the system’s final temperature will higher than 0°C. Assume that the final temperature is T of the punch plus ice system. )0()()( 321 ??+=?+=++ TCmmTCmmQQQ watericewaterwatericewater )K(53.281)C(38.8 )45.0(4168 106675.110025.1103488.3 )( 545 321 == +× ×?×?× = + ++ ?= o watericewater Cmm QQQ T 2. A typical household water heater has a capacity of about 0.50 m 3 of water. The water is initially at a temperature of 15 °C and is warmed to 60 °C for the pleasure of long, hot showers. The quantity of the heat transfer which is necessary to warm the water is 9.42×10 7 J . If the heating element is rated at 5.0 kW, the time it take to warm the water is 5.2 h . Solution: (a) The quantity of the heat transfer which is necessary to warm the water is )J(1042.9)1560(41865.0101 73 ×=?××××= ?=?= TCVTCmQ waterwaterwaterwaterwater ρ (b) tPQ ?=Q )h(2.5)s(1084.18 105 1042.9 3 3 7 =×= × × ==?∴ P Q t 3. Two reservoirs, one at 0.0 °C and the other at 100.0 °C, are connected by means of two materials as indicated in Figure 2. The relative R-values of the materials are specified. The temperature at the interface between the two materials is 348.15 K . Solution: Assume that the temperature is T at the interface between the two materials. For R TT A t Q R TT A t Q CH 3 ) d d (,) d d ( 21 ? = ? = and 21 ) d d () d d ( t Q t Q = So we have )K(15.348 4 15.273)10015.273(3 4 3 3 3 3 = ++× = + = + + =? ? = ? CHCH CH TT RR RTRT T R TT A R TT A 4. A gas is taken from thermal equilibrium state 1 to state 2 on a P-V diagram via the semicircular route indicated in Figure 3. The work done by the gas in going from state 1 to state 2 is 2.61×10 3 J . The work done by the gas if it goes from state 2 back to state 1, retracting the semicircular path is -2.61×10 3 J . Solution: (a) The work is equal to the area under the semicircular route )J(1061.2)10 2 110 10013.14( 2 10)110(10013.16 33535 ×=× ? ×××?×?×××= ?? π W (b) )J(1061.2 3 ×?=?=′ WW 5. Gas within a chamber passes through the cycle shown in Fig.4. The net heat added to the gas during process CA if Q AB =20J, 0= BC Q and W BCA = 15J is -5J . Solution: V P A B C Fig.4 0 V(liters) P(atm) 0 Fig.3 1.00 10.00 2.00 6.00 Hot reservior Cold reservior 3R R Fig.2 CACACABCBCBCABAB WUQWUQUQ +?=+?=?= ,, For the cycle, 0=?+?+?=? CABCAB UUUU , Therefore, BCACABCAB WQQQ =++ J52015 ?=?=?= ABBCACA QWQ 6. Gas within a chamber undergoes the processes shown in the PV diagram of Fig. 5. The net heat added to the system during one complete cycle is -3.58×10 3 J . Solution: The work during one complete cycle is equal to the area of the semicircular route on the diagram: )J(1058.3]10 2 14 10013.1)1530[( 2 335 ×?=× ? ×××??= ? π W Applying the first law of thermodynamics: WUQ +?= and 0=?U , so the net heat added to the system during one complete cycle is J1058.3 3 ×?==WQ III. Give the Solutions of the Following Problems 1. When a system is taken from state i to state f along the path iaf in Fig.6, it is found that Q=50J and W=20J. Along the path ibf, Q=36J. (a) What is W along the path ibf? (b) If W=-13J for the curved return path fi, what is Q for this path? (c) Take E int,i =10J. What is E int,f ? (d) If E int,b =22J, find Q for process ib and prosess bf. Solution: (a) Applying the first law of thermodynamics: WUQ +?= , so WQUUU if ?=?=? , that is ibfibfiafiaf WQWQ ?=? Thus )J(6502036 =?+=?+= iafiafibfibf QWQW (b) The same way as (a), we have fifiiafiaf WQWQ ?=? So )J(17132050 =??=+?= fiiafiaffi WWQQ (c) Applying the first law of thermodynamics: WUQ +?= , so V(L) P（Mpa） 0 Fig.5 1 2 3 4 5 10 20 30 40 V P Fig.6 i f a b )J(40102050 intintintint =+?=+?=??=?=? iiafiaffiafiafif EWQEWQEEU (d) See the diagram, J6== ibibf WW . Applying the first law of thermodynamics, the heat transfer for process ib is )J(1861022 intint =+?=+?=+?= ibib WEEWUQ As bfibibf QQQ += , so the heat transfer for process bf is )J(181836 =?=?= ibibfbf QQQ 2. Three (3.00) moles of an ideal gas is taken along the excursion on a P-V diagram indicated in Figure 7. The first part of the path is an isobaric process and the second part of the path is isothermal. (a) What is the initial absolute temperature of the gas? (b) Calculate the work done by the gas along the isobaric segment of the path. (c) What is the temperature of the gas along the isotherm? (d) Is the work done by the gas positive or negative along the isothermal segment of the path? (e) Determine the final volume V f of the gas so that the total work done by the gas is zero. Express your result in liters. (f)What is the final pressure P f of the gas? Express your result in atmospheres. Solution: (a) Applying the equation of ideal gas nRTPV = , the initial absolute temperature of the gas is )K(51.300 3315.8 10410013.15.18 35 = × ×××× == ? Rn PV T (b) The work done by the gas along the isobaric segment of the path is )J(05.187410)45(10013.15.18 35 =×?×××=?= ? VPW iP (c) Applying the equation of ideal gas nRTPV = , the temperature of the gas along the isotherm is )K(64.375 3315.8 10510013.15.18 35 = × ×××× == ? Rn VP T i f (d) The work is negative along the isothermal segment of the path. (e) In order that the total work done by the gas is zero, we have A V(liters) P(atm) Fig.7 4.0 5.00 18.5 P f B C P i f fT W V V nRTW ?== ln So the final volume V f of the gas is )l(09.4)m(1009.4105 33 64.3753315.8 05.1874 3 =×=××== ? ×× ? ? ? eeVV f P nRT W if (f) Applying the equation of ideal gas nRTPV = , the final pressure P f of the gas is )atm(62.22)Pa(1029.2 1009.4 64.3753315.8 6 3 =×= × ×× == ? f f f V nRT P 3. Suppose that a sample of gas expands from 2.0 to 8.0 m 3 along the diagonal path in the PV diagram shown in Fig.8. It is then compressed back to 2.0 m 3 along either path 1 or path 2. Compute the net work done on the gas for the complete cycle in each case. Solution: The net work done on the gas is the area of the cycle on the diagram. The area of the cycle along path 1 is 43 1 105.4)82(10)520( 2 1 ×?=?××?×=S So the net work done on the gas along path 1 is )J(105.4 4 1 ×?=W The area of the cycle along path 2 is 43 2 105.4)28(10)520( 2 1 ×=?××?×=S So the net work done on the gas along path 2 is )J(105.4 4 1 ×=W 4. A gas is taken around a closed cycle on a P-V diagram as indicated in Figure 9. (a) Find the total work W done by the gas. (b) What is the change in the internal energy of the gas? (c) What is the total heat transfer to the gas in executing this cycle? If the gas is taken around the cycle in the opposite V(liters) P(atm) 0 Fig.9 3.00 6.00 2.00 6.00 V(m 3 ) P（kpa） 0 Fig.8 2 4 6 8 5 10 15 20 Path 1 Path 2 direction: (d) Find the total work done by the gas; (e) Find the change in the internal energy of the gas; (f) Find the total heat transfer to the gas in executing the cycle in this direction. Solution: (a) The total work W done by the gas is )J(10078.610)36(10013.1)26( 2 1 235 ×=×?×××?== ? ? SW . (b) The change in the internal energy of the gas is )J(0=?U . (c) Applying the first law of thermodynamics: WUQ +?= and 0=?U , the total heat transfer to the gas in executing this cycle is )J(10078.6 2 ×==WQ . (d) The total work W done by the gas is )J(10078.6 2 ×??=′ WW . (e) The change in the internal energy of the gas is )J(0=?U . (f) The total heat transfer to the gas in executing the cycle in opposite direction is )J(10078.6 2 ×?==WQ .