Molar mass: NAm?6.02?1023?5.31?10?26?32.0g, oxygen gas. (c) ??nm?0.96kg/m3. (d)
55kTn??1.38?10?23?400?1.8?1025?2.5?105J/m3. 22
Problem 8. Answers: 1. 0.92c L?L01?v2/c2, v?c1?22/52?0.92c 2. (a) 2.2?10?6 s, (b) 653 m.
L Solution: ?t?, ?t0??t1?v2/c2 v4.6?103?t0?1?0.992?2.2?10?6s,
0.99c ?x0?v?t0?653m
3. b. (the constancy of the speed of light) 4. b. As the figure shows
VAH1??1?v2/c2?0.2? V0AH05
5. (a) L0 = 17.4m , (b) ? = 3.3?
Solution:
(a)l0x?lx/1?v2/c2?lcos30?/1?0.9952 l0y?ly?lsin30?,l0?l02x?l02y?17.4m Solution: (b)
Fig. 8-5
?tan?0?l0yl0x?tan301?0.995, ?0?3.3
?26. (a) 2.50?108 m/s, (b) 4.97 m, (c) ? 1.33?10?8 s. Solution: (a)
From the given condition, ?x??0
From the Lorentz transformation, ?x???(?x?v?t)?0
v??x5?38??2.50?10m/s ?9?t(9?1)?10Solution: (b) From the Lorentz transformation
t , x???(x?v)??xB???(xR?vtR)? xR11?(2.5/3)2(3?2.5?108?10?9)?4.97m,
Solution (c) From the Lorentz transformation
vxvx???(tR?2R)??1.33?10?8s t???(t?2), tRccvx???(tB?2B)??4.98?10?9s. In S?, red flash occurs first. tBcProblem 11. Answers: 1. (a) 3.5?103 J, (b) 2.5?103 J , (c) ?1000 J.
77 Solution: Q?Cp,m?T?R?T??8.31?120?3.5?103J
2255 ?E?CV,m?T?R?T??8.31?120?2.5?103J
22 Work done by the gas:
W?Q??E?1000J
Work done on the gas: W??1000J 2. 227K,
??1Solutio: TV?T2V2??1, (11V2??1T1)?, V1T2(V2??1T1T1300)?,T2?0.4?0.4?227K V1T222
3. b . As the figure shows.
4. e.
Solution: For a free expansion:
?E?0,W?0,Q?0, pV11?p2V2
V2?V1,p1?p2
5. Solution: (a) paVa?RTa,pcVc?RTc
?E?CV?T?5RpcVc?paVa5?(4?2?5?2)?103??5kJ
2R2
(b)
Q??E?W, W?1(2?5)?2?103?7kJ 2 ?Q??E?W?5?7?2?k. J (c) ?E??5kJ, W?5?2?103?10kJ
E?W?5?1?0 ?Q??5?k J
Fig. 11-5
6. Solution: From the definition for molar specific heat at constant pressure, we
have
Cp?(dQ)p , dTFrom the first Law of thermal dynamics, we have
(dQ)p?dE?dW?CVdT?pdV, Here, dE?CVdT Combining theses equations, obtains
CdT?pdVdQ)p?V dTdTpdV Cp?CV?,
dT Cp?(From the ideal gas law, for the constant pressure process with one mole of ideal gas, we have
pdV?Rd T
RdT dTThen Cp?CV?Therefore Cp?CV?R
Problem 12. Answer: 1. 3.28 J/K
Solution: The entropy change of the Universe due to the energy
transfer by radiation from the Sun to the Earth is
?Suniverse??Ssun??Searth?103103???3.28J/K. 58002902. 57.2 J/K.
Solution: Suppose the process can be replaced by a reversible isothermal
mgh?57.2J. /Kprocess, then ?Sai?rla?keT3. d. ?S??Sgas??Senr?0
4. c.
Solution: free expansion is an irreversible process which occurs in an isolated
system. ?S?0
5. (a) ?0.390?R, (b) ?0.545?R,
273?25?CdT5298V???Rln??0.390?R Solution: (a) ?S???273?18T2255 (b) ?S???273?25?CpdTT273?18???7298Rln??0.545?R 22556. (a) 4500 J, (b) ? 4986 J, (c) 9486 J
dQSolution: (a) dS?
TdQ?TdS, Q??TdS?The area under the T?S curve
