中南大学大学物理双语版答案Problem 1-22

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