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《大学物理简明教程》答案[1]

《大学物理简明教程》答案[1]
《大学物理简明教程》答案[1]

大学物理简明教程习题解答

习题一

1-1 |r ?|与r ?有无不同?t d d r 和t d d r 有无不同? t d d v 和t d d v

有无不同?其不同在哪里?

试举例说明.

解:(1)r ?是位移的模,?r 是位矢的模的增量,即

r ?12r r -=,12r r r -=?; (2)t d d r 是速度的模,即t d d r ==v t

s

d d .

t r

d d 只是速度在径向上的分量.

∵有r r ?r =(式中r ?叫做单位矢),则t ?r ?t r t d d d d d d r r

r += 式中t r

d d 就是速度径向上的分量,

t r t d d d d 与r 不同如题1-1图所示.

题1-1图

(3)t d d v 表示加速度的模,即

t v a d d

=

,t v d d 是加速度a 在切向上的分量. ∵有ττ (v =v 表轨道节线方向单位矢),所以

t v t v t v d d d d d d ττ +=

式中dt dv

就是加速度的切向分量.

(t t

r d ?d d ?d τ 与

的运算较复杂,超出教材规定,故不予讨论) 1-2 设质点的运动方程为x =x (t ),y =y (t ),在计算质点的速度和加速度时,有人先求

出r =2

2y x +,然后根据v =t r

d d ,及a =22d d t r 而求得结果;又有人先计算速度和加速度

的分量,再合成求得结果,即

=2

2

d d d d ??? ??+??? ??t y t x 及a =

2

22222d d d d ?

??? ??+???? ??t y t x 你认为两种方法哪一种正确?为什么?两者差别何在?

解:后一种方法正确.因为速度与加速度都是矢量,在平面直角坐标系中,有j y i x r

+=,

j

t y i t x t r a j

t y i t x t r v

222222d d d d d d d d d d d d +==+==∴

故它们的模即为

2

222

222

22

22

2d d d d d d d d ?

??? ??+???? ??=+=?

?

? ??+??? ??=+=t y t x a a a t y t x v v v y x y x

而前一种方法的错误可能有两点,其一是概念上的错误,即误把速度、加速度定义作

2

2d d d d t r a t

r

v ==

其二,可能是将2

2d d d d t r t

r 与误作速度与加速度的模。在1-1题中已说明t r d d 不是速度的模,而只是速度在径向上的分量,同样,2

2d d t r 也不是加速度的模,它只是加速度在径向分量中

的一部分????

???????

??-=2

22d d d d t r t r a θ径。或者概括性地说,前一种方法只考虑了位矢r 在径向(即

量值)方面随时间的变化率,而没有考虑位矢r

及速度v 的方向随间的变化率对速度、加速

度的贡献。

1-3 一质点在xOy 平面上运动,运动方程为

x =3t +5, y =21

t 2+3t -4.

式中t 以 s 计,x ,y 以m 计.(1)以时间t 为变量,写出质点位置矢量的表示式;(2)求出t =1 s 时刻和t =2s 时刻的位置矢量,计算这1秒内质点的位移;(3)计算t =0 s 时刻到t =4s 时刻内的平均速度;(4)求出质点速度矢量表示式,计算t =4 s 时质点的速度;(5)计算t =0s 到t =4s 内质点的平均加速度;(6)求出质点加速度矢量的表示式,计算t =4s 时质点的加速度(请把位置矢量、位移、平均速度、瞬时速度、平均加速度、瞬时加速度都表示成直角坐标系中的矢量式).

解:(1)

j

t t i t r

)4321()53(2-+++=m (2)将1=t ,2=t 代入上式即有

j i r

5.081-= m

j j r

4112+=m

j j r r r

5.4312+=-=?m

(3)∵ j i r j j r 1617,4540

+=-= ∴ 1

04s m 534201204-?+=+=--=??=j i j i r r t r v

(4) 1

s m )3(3d d -?++==j t i t r v

则 j i v 734

+= 1s m -? (5)∵ j i v j i v

73,3340

+=+=

2

04s m 1444-?==-=??=j v v t v a (6) 2

s m 1d d -?==j t v

a

这说明该点只有y 方向的加速度,且为恒量。

1-4 在离水面高h 米的岸上,有人用绳子拉船靠岸,船在离岸S 处,如题1-4图所示.当人以

0v (m ·1-s )

的速率收绳时,试求船运动的速度和加速度的大小.

图1-4

解: 设人到船之间绳的长度为l ,此时绳与水面成θ角,由图可知

2

22s h l += 将上式对时间t 求导,得

t s s

t

l l

d d 2d d 2=

题1-4图

根据速度的定义,并注意到l ,s 是随t 减少的,

t s v v t l v d d ,d d 0-==-

=船绳

θcos d d d d 00v v s l t l s l t s v ==-=-

=船 或 s v s h s lv v 0

2/1220)(+=

=船

将船v 再对t 求导,即得船的加速度

3

2

0222

020

2

002)(d d d d d d s v h s v s l s v s

lv s v v s t s

l t l s

t v a =+-=+-=-==船

1-5 质点沿x 轴运动,其加速度和位置的关系为 a =2+62x ,a 的单位为2

s m -?,x 的单位

为 m. 质点在x =0处,速度为101

s m -?,试求质点在任何坐标处的速度值.

解: ∵

x v v t x x v t v a d d d d d d d d ===

分离变量:

x x adx d )62(d 2

+==υυ 两边积分得 c

x x v ++=32

2221

由题知,0=x 时,100

=v ,∴50=c

∴ 1

3s m 252-?++=x x v

1-6 已知一质点作直线运动,其加速度为 a =4+3t 2

s m -?,开始运动时,x =5 m , v =0,求该质点在t =10s 时的速度和位置.

解:∵ t t v

a 34d d +==

分离变量,得 t t v d )34(d +=

积分,得 1

223

4c t t v ++=

由题知,0=t ,00

=v ,∴01=c

故 2234t t v +

=

又因为

2

234d d t t t x v +== 分离变量, t

t t x d )23

4(d 2+=

积分得 2

3221

2c t t x ++=

由题知 0=t ,50

=x ,∴52=c

故 52123

2++

=t t x

所以s 10=t 时

m

70551021

102s m 1901023

10432101210=+?+?=?=?+

?=-x v

1-7 一质点沿半径为1 m 的圆周运动,运动方程为 θ=2+33

t ,θ式中以弧度计,t 以秒

计,求:(1) t =2 s 时,质点的切向和法向加速度;(2)当加速度的方向和半径成45°角时,其角位移是多少?

解:

t t t t 18d d ,9d d 2====

ωβθω

(1)s 2=t 时, 2

s m 362181-?=??==βτR a

2222s m 1296)29(1-?=??==ωR a n

(2)当加速度方向与半径成ο45角时,有1

45tan ==?n

a a

τ

βωR R =2

亦即

t t 18)9(2

2= 则解得 923=

t 于是角位移为

rad

67.292

32323=?+=+=t θ 1-8 质点沿半径为R 的圆周按s =2

021bt t v -的规律运动,式中s 为质点离圆周上某点的弧

长,

0v ,b 都是常量,

求:(1)t 时刻质点的加速度;(2) t 为何值时,加速度在数值上等于b .

解:(1) bt

v t s v -==0d d R bt v R v a b

t

v a n 2

02)(d d -=

=-==τ

则 24

02

2

2

)(R bt v b a a a n

-+

=+=τ

加速度与半径的夹角为

20)(arctan

bt v Rb a a n --=

=τ?

(2)由题意应有

24

02

)(R bt v b b a -+

== 即 0

)(,)

(4024

022=-?-+=bt v R bt v b b

∴当

b v t 0

=时,b a = 1-9 以初速度

0v =201s m -?抛出一小球,抛出方向与水平面成幔60°的夹角,

求:(1)球轨道最高点的曲率半径1R ;(2)落地处的曲率半径2

R .

(提示:利用曲率半径与法向加速度之间的关系)

解:设小球所作抛物线轨道如题1-10图所示.

题1-9图

(1)在最高点,

o 0160cos v v v x == 2

1s m 10-?==g a n

又∵

121

1ρv

a n =

m

1010

)60cos 20(2

2111=??=

=n a v ρ

(2)在落地点,

2002==v v 1s m -?,

o

60cos 2?=g a n

∴ m

8060cos 10)20(2

2222

=??==n a v ρ

1-10飞轮半径为0.4 m ,自静止启动,其角加速度为 β=0.2 rad ·2

s -,求t =2s 时边缘上各点的速度、法向加速度、切向加速度和合加速度.

解:当s 2=t 时,4.022.0=?==t βω1

s rad -? 则16.04.04.0=?==ωR v 1

s m -?

064.0)4.0(4.022=?==ωR a n 2s m -? 08.02.04.0=?==βτR a 2s m -?

2

2222

s m 102.0)08.0()064.0(-?=+=+=τa a a n

1-11 一船以速率1v =30km ·h -1

沿直线向东行驶,另一小艇在其前方以速率2v =40km ·h -1

沿直线向北行驶,问在船上看小艇的速度为何?在艇上看船的速度又为何?

解:(1)大船看小艇,则有1221

v v v -=,依题意作速度矢量图如题1-13图(a)

题1-11图

由图可知 1222121h km 50-?=+=v v v

方向北偏西

?===87.3643

arctan arctan

21v v θ

(2)小船看大船,则有2112

v v v -=,依题意作出速度矢量图如题1-13图(b),同上法,得 5012=v 1h km -?

方向南偏东o

87

.36

习题二

2-1 一个质量为P 的质点,在光滑的固定斜面(倾角为α)上以初速度

0v 运动,0v 的方向

与斜面底边的水平线AB 平行,如图所示,求这质点的运动轨道.

解: 物体置于斜面上受到重力mg ,斜面支持力N .建立坐标:取0v

方向为X 轴,平行斜

面与X 轴垂直方向为Y 轴.

如图2-2.

题2-1图

X 方向: 0=x F t v x 0= ① Y 方向: y y ma mg F ==αsin ②

0=t 时 0=y 0=y v

2

sin 21

t g y α=

由①、②式消去t ,得

220

sin 21x g v y ?=

α

2-2 质量为16 kg 的质点在xOy 平面内运动,受一恒力作用,力的分量为x f =6

N ,y f =

-7 N ,当t =0时,==y x 0,x v

=-2 m ·s -1

,y

v =0.求

当t =2 s 时质点的 (1)位矢;(2)速度.

解:

2s m 83166-?===

m f a x x 2

s m 167-?-==m f a y y

(1)

??--?-=?-=+=?-=?+-=+=201

01

200s m 87

2167s m 45

2832dt a v v dt a v v y y y x x x

于是质点在s 2时的速度

1

s m 8

745-?--=j

i v

(2)

m

874134)16

7(21)483

2122(2

1)21(220j i j

i j

t a i t a t v r y x

--=?-+??+?-=++=

2-3 质点在流体中作直线运动,受与速度成正比的阻力kv (k 为常数)作用,t =0时质点的

速度为

0v ,证明(1) t 时刻的速度为v =t m

k e

v )(0-;(2) 由0到t 的时间内经过的距离为

x =(k m v 0)[1-t m

k

e )(-];(3)停止运动前经过的距离为

)(0k m v ;(4)证明当k m t =时速度减至0v 的e 1,式中m 为质点的质量.

答: (1)∵

t v m kv a d d =

-= 分离变量,得

m t k v v d d -= 即 ??-=v v t m t

k v

v 00d d

m

kt e v v -=ln ln 0

t

m k

e

v v -=0

(2)

??---===t

t

t

m k m k e k mv t e

v t v x 0

00)

1(d d

(3)质点停止运动时速度为零,即t →∞,

故有

?∞

-=

='0

0d k mv t e

v x t

m k

(4)当t=k m

时,其速度为

e v e v e

v v k

m

m k 0

100=

==-?-

即速度减至0v 的e 1.

2-4一质量为m 的质点以与地的仰角θ=30°的初速0v 从地面抛出,若忽略空气阻力,求质

点落地时相对抛射时的动量的增量.

解: 依题意作出示意图如题2-6图

题2-4图

在忽略空气阻力情况下,抛体落地瞬时的末速度大小与初速度大小相同,与轨道相切斜向下, 而抛物线具有对y 轴对称性,故末速度与x 轴夹角亦为o

30,则动量的增量为

0v m v m p -=?

由矢量图知,动量增量大小为

v m

,方向竖直向下.

2-5 作用在质量为10 kg 的物体上的力为i t F

)210(+=N ,式中t 的单位是s ,(1)求4s 后,

这物体的动量和速度的变化,以及力给予物体的冲量.(2)为了使这力的冲量为200 N ·s ,

该力应在这物体上作用多久,试就一原来静止的物体和一个具有初速度j

6-m ·s -1

的物体,

回答这两个问题.

解: (1)若物体原来静止,则

i

t i t t F p t

10401s m kg 56d )210(d -??=+==???,沿x 轴正向,

i p I i

m p v

11111

1s m kg 56s m 6.5--??=?=?=?=? 若物体原来具有6-1

s m -?初速,则

??+-=+-=-=t t t

F v m t m F v m p v m p 0

00000d )d (,

于是 ??==-=?t p t F p p p 01

02d

, 同理, 12

v v

?=?,12I I = 这说明,只要力函数不变,作用时间相同,则不管物体有无初动量,也不管初动量有多大,

那么物体获得的动量的增量(亦即冲量)就一定相同,这就是动量定理. (2)同上理,两种情况中的作用时间相同,即

?+=+=t

t t t t I 02

10d )210(

亦即 0200102

=-+t t 解得s 10=t ,(s 20='t 舍去)

2-6 一颗子弹由枪口射出时速率为1

0s m -?v ,当子弹在枪筒内被加速时,它所受的合力为

F =(bt a -)N(b a ,为常数),其中t 以秒为单位:(1)假设子弹运行到枪口处合力刚好为零,

试计算子弹走完枪筒全长所需时间;(2)求子弹所受的冲量.(3)求子弹的质量.

解: (1)由题意,子弹到枪口时,有

0)(=-=bt a F ,得

b a t =

(2)子弹所受的冲量

?-=-=t bt at t bt a I 02

21d )(

b a

t =

代入,得

b a I 22=

(3)由动量定理可求得子弹的质量

02

02bv a v I m =

=

2-7设N 67j i F -=合.(1) 当一质点从原点运动到m 1643k j i r ++-=时,求F 所作的功.(2)如果质点到r 处时需0.6s ,试求平均功率.(3)如果质点的质量为1kg ,试求动能的

变化.

解: (1)由题知,合F 为恒力,

∴ )1643()67(k j i j i r F A

++-?-=?=合 J 452421-=--=

(2) w 756.045==?=

t A P

(3)由动能定理,

J 45-==?A E k

2-8 如题2-18图所示,一物体质量为2kg ,以初速度0v

=3m ·s -1

从斜面A 点处下滑,它与

斜面的摩擦力为8N ,到达B 点后压缩弹簧20cm 后停止,然后又被弹回,求弹簧的劲度系数和物体最后能回到的高度.

解: 取木块压缩弹簧至最短处的位置为重力势能零点,弹簧原 长处为弹性势能零点。则由功能原理,有

??? ???+-=

-37sin 212122mgs mv kx s f r 22

2137sin 2

1kx s f mgs m v k r -?+=

式中m 52.08.4=+=s ,m 2.0=x ,再代入有关数据,解得

-1m N 1390?=k

题2-8图

再次运用功能原理,求木块弹回的高度h '

2o 2137sin kx s mg s f r -

'='-

代入有关数据,得 m 4.1='s , 则木块弹回高度

m 84.037sin o ='='s h

2-9 一个小球与一质量相等的静止小球发生非对心弹性碰撞,试证碰后两小球的运动方向

互相垂直.

证: 两小球碰撞过程中,机械能守恒,有

222120212121mv mv mv +=

即 2

22120v v v +=

题2-9图(a) 题2-9图(b) 又碰撞过程中,动量守恒,即有

210v m v m v m +=

亦即 210

v v v += ②

由②可作出矢量三角形如图(b),又由①式可知三矢量之间满足勾股定理,且以0v 为斜边,

故知1v 与2v 是互相垂直的.

2-10一质量为

m 的质点位于(

11,y x )处,速度为

j v i v v y x

+=, 质点受到一个沿x

负方向

的力f 的作用,求相对于坐标原点的角动量以及作用于质点上的力的力矩.

解: 由题知,质点的位矢为

j y i x r

11+= 作用在质点上的力为 i f f -=

所以,质点对原点的角动量为

v m r L ?=0

)

()(11j v i v m i y i x y x +?+=

k

mv y mv x x y

)(11-=

作用在质点上的力的力矩为

k f y i f j y i x f r M

1110)()(=-?+=?=

2-11 哈雷彗星绕太阳运动的轨道是一个椭圆.它离太阳最近距离为1r =8.75×1010

m 时的速

率是1v =5.46×104

m ·s -1

,它离太阳最远时的速率是2v =9.08×102

m ·s -1

这时它离太

阳的距离2r 多少?(太阳位于椭圆的一个焦点。)

解: 哈雷彗星绕太阳运动时受到太阳的引力——即有心力的作用,所以角动量守恒;又由于哈雷彗星在近日点及远日点时的速度都与轨道半径垂直,故有

2211mv r mv

r = ∴ m 1026.51008.91046.51075.812

2

4102112?=????==v v r r

2-12 物体质量为3kg ,t =0时位于m 4i r

=, 1s m 6-?+=j i v ,如一恒力N 5j f =作用在物体上,求3秒后,(1)物体动量的变化;(2)相对z 轴角动量的变化.

解: (1)

??-??===?30

1

s m kg 15d 5d j t j t f p

(2)解(一)

73400=+=+=t v x x x

j

at t v y y 5.25335

213621220=??+?=+= 即 i r

41=,j i r 5.2572+=

10==x x v v

11

335

60=?+=+=at v v y y 即 j i v

611+=,j i v 112+=

∴ k j i i v m r L

72)6(34111=+?=?=

k j i j i v m r L

5.154)11(3)5.257(222=+?+=?=

∴ 1

212s m kg 5.82-??=-=?k L L L

解(二) ∵

dt dz

M =

???=?=?t t t

F r t M L 0

d )(d

??-??=+=???

?????+++=3

1

302s m kg 5.82d )4(5d 5)35)216()4(2k t k t t j j t t i t

题2-12图

2-13飞轮的质量m =60kg ,半径R =0.25m ,绕其水平中心轴O 转动,转速为

900rev ·min -1

.现利用一制动的闸杆,在闸杆的一端加一竖直方向的制动力F ,可使飞轮减速.已知闸杆的尺寸如题2-25图所示,闸瓦与飞轮之间的摩擦系数μ=0.4,飞轮的转动惯量可按匀质圆盘计算.试求:

(1)设F =100 N ,问可使飞轮在多长时间内停止转动?在这段时间里飞轮转了几转? (2)如果在2s 内飞轮转速减少一半,需加多大的力F ?

解: (1)先作闸杆和飞轮的受力分析图(如图(b)).图中N 、N '是正压力,r F 、r F '是摩擦力,

x F 和y F 是杆在A 点转轴处所受支承力,R 是轮的重力,P 是轮在O

轴处所受支承力.

题2-13图(a

题2-13图(b)

杆处于静止状态,所以对A 点的合力矩应为零,设闸瓦厚度不计,则有

F l l l N l N l l F 1

2

11210

)(+=

'='-+

对飞轮,按转动定律有I R F r /-=β,式中负号表示β与角速度ω方向相反. ∵ N F r μ= N N '= ∴

F l l l N F r 1

2

1+='=μ

μ

又∵

,21

2mR I =

F mRl l l I R F r 1

21)(2+-=-

=μβ ①

以N 100=F 等代入上式,得

2

s rad 340

10050.025.060)75.050.0(40.02-?-=???+??-=

β

由此可算出自施加制动闸开始到飞轮停止转动的时间为

s 06.740603

29000=???=-

=πβωt

这段时间内飞轮的角位移为

rad 21.53)4

9

(3402149602900212

20ππππβωφ?=??-??=

+=t t

可知在这段时间里,飞轮转了1.53转.

(2)1

0s rad 602900-??=π

ω,要求飞轮转速在2=t s 内减少一半,可知

20

00

s rad 21522

-?-

=-

=-=π

ωωωβt

t

用上面式(1)所示的关系,可求出所需的制动力为

N l l m Rl F 1772)75.050.0(40.021550.025.060)

(2211=?+?????=

+-=π

μβ

2-14固定在一起的两个同轴均匀圆柱体可绕其光滑的水平对称轴O O '转动.设大小圆柱体的半径分别为R 和r ,质量分别为M 和m .绕在两柱体上的细绳分别与物体1m 和2m 相连,

1m 和2m 则挂在圆柱体的两侧,如题2-26图所示.设R =0.20m, r =0.10m ,m =4 kg ,

M =10 kg ,1m =2m =2 kg ,且开始时1m ,2m 离地均为h =2m .求:

(1)柱体转动时的角加速度;

(2)两侧细绳的张力.

解: 设1a ,2a 和β分别为1m ,2m 和柱体的加速度及角加速度,方向如图(如图b).

题2-14(a)图 题2-14(b)图 (1) 1m ,2m 和柱体的运动方程如下:

2222a m g m T =- ① 1111a m T g m =- ②

βI r T R T ='

-'21 ③

式中 ββR a r a T T T T ==='='122211,,, 而 222121mr MR I +=

由上式求得

2

22222

2212

1s rad 13.68

.910.0220.0210.0421

20.010212

1.02

2.0-?=??+?+??+???-?=

++-=

g

r m R m I rm Rm β

(2)由①式

8.208.9213.610.02222=?+??=+=g m r m T βN

由②式

1.1713.6.

2.028.92111=??-?=-=βR m g m T N

2-15 如题2-15图所示,一匀质细杆质量为m ,长为l ,可绕过一端O 的水平轴自由转动,

杆于水平位置由静止开始摆下.求: (1)初始时刻的角加速度; (2)杆转过θ角时的角速度. 解: (1)由转动定律,有

β)31

(212ml mg

=

l g 23=

β (2)由机械能守恒定律,有

22)31

(21sin 2ωθml l mg

=

l g θωsin 3=

题2-15图

习题四

4-1下列表述是否正确?为什么?并将错误更正. (1)A E Q ?+?=? (2)

?+=V

p E Q d

(3)

121Q Q -

≠η (4)12

1Q Q

-<不可逆η 解:(1)不正确,A E Q +?=

(2)不正确,

?+=V

p E Q d Δ

(3)不正确,

121Q Q -

(4)不正确,

12

1Q Q -=不可逆

η

4-2 用热力学第一定律和第二定律分别证明,在V p -图上一绝热线与一等温线不能有两个交点.

题4-2图

解:1.由热力学第一定律有

A E Q +?= 若有两个交点a 和b ,则 经等温b a →过程有

0111=-=?A Q E 经绝热b a →过程

012=+?A E 022<-=?A E

从上得出21E E ?≠?,这与a ,b 两点的内能变化应该相同矛盾.

2.若两条曲线有两个交点,则组成闭合曲线而构成了一循环过程,这循环过程只有吸热,无放热,且对外做正功,热机效率为%100,违背了热力学第二定律. 4-3 一循环过程如题4-3图所示,试指出: (1)ca bc ab ,,各是什么过程;

(2)画出对应的V p -图; (3)该循环是否是正循环?

(4)该循环作的功是否等于直角三角形面积? (5)用图中的热量

ac bc ab Q Q Q ,,表述其热机效率或致冷系数.

解:(1) a b 是等体过程

bc 过程:从图知有KT V =,K 为斜率 由vRT pV = 得 K vR p =

故bc 过程为等压过程 ca 是等温过程 (2)V p -图如题4-3’图

题4-3’图

(3)该循环是逆循环

(4)该循环作的功不等于直角三角形面积,因为直角三角形不是V p -图中的图形.

(5)

ab ca bc ab

Q Q Q Q e -+=

题4-3图 题4-4图

4-4 两个卡诺循环如题4-4图所示,它们的循环面积相等,试问: (1)它们吸热和放热的差值是否相同; (2)对外作的净功是否相等; (3)效率是否相同?

答:由于卡诺循环曲线所包围的面积相等,系统对外所作的净功相等,也就是吸热和放热的差值相等.但吸热和放热的多少不一定相等,效率也就不相同. 4-5 根据

?

=-B

A

A B T

Q S S 可逆

d 及

?

>-B

A A

B T Q S S 不可逆d ,这是否说明可逆过程的熵变大于

不可逆过程熵变?为什么?说明理由.

答:这不能说明可逆过程的熵变大于不可逆过程熵变,熵是状态函数,熵变只与初末状态有关,如果可逆过程和不可逆过程初末状态相同,具有相同的熵变.只能说在不可逆过程中,系统的热温比之和小于熵变.

4-6 如题4-6图所示,一系统由状态a 沿acb 到达状态b 的过程中,有350 J 热量传入系统,而系统作功126 J .

(1)若沿adb 时,系统作功42 J ,问有多少热量传入系统?

(2)若系统由状态b 沿曲线ba 返回状态a 时,外界对系统作功为84 J ,试问系统是吸热还是放热?热量传递是多少?

题4-6图

解:由abc 过程可求出b 态和a 态的内能之差 A E Q +?= 224126350=-=-=?A Q E J abd 过程,系统作功42=A J

26642224=+=+?=A E Q J 系统吸收热量

ba 过程,外界对系统作功84-=A J

30884224-=--=+?=A E Q J 系统放热

4-7 1 mol 单原子理想气体从300 K 加热到350 K ,问在下列两过程中吸收了多少热量?增加了多少内能?对外作了多少功? (1)体积保持不变; (2)压力保持不变. 解:(1)等体过程

由热力学第一定律得E Q ?= 吸热

)(2)(1212V T T R i

T T C E Q -=-=?=υ

υ

25.623)300350(31.823

=-??=

?=E Q J

对外作功 0=A (2)等压过程

)(22

)(1212P T T R i T T C Q -+=-=υ

υ

吸热 75

.1038)300350(31.825

=-??=Q J

)(12V T T C E -=?υ

内能增加 25.623)300350(31.823

=-??=

?E J

对外作功 5.4155.62375.1038

=-=?-=E Q A J 4-8 0.01 m 3

氮气在温度为300 K 时,由0.1 MPa(即1 atm)压缩到10 MPa .试分别求氮

气经等温及绝热压缩后的(1)体积;(2)温度;(3)各过程对外所作的功. 解:(1)等温压缩 300=T K 由2211V p V p = 求得体积

3

211210101.0101

-?=?==

p V

p V 3m

对外作功

21112ln ln

p p

V p V V VRT A == 01.0ln 01.010013.115

????= 31067.4?-=J

(2)绝热压缩

R C 25

V = 57=

γ 由绝热方程 γ

γ2211V p V p = γ

γ/12112)(p V p V =

1

1

2

1/12112)()(V p p

p V p V γγγ==

3

411093.101.0)101

(-?=?=m

由绝热方程γ

γγγ---=22111p T p T 得

K 579)10(30024

.04.11

1

1

212=?==--T p p T T γγγγ

热力学第一定律A E Q +?=,0=Q 所以

)(12mol

T T C M M

A V --

=

RT M M

pV mol

=

,)

(2512111T T R RT V p A --=

3

5105.23)300579(25300001.010013.1?-=-????-=A J

4-9 1 mol 的理想气体的T-V 图如题4-9图所示,ab 为直线,延长线通过原点O .求ab 过

程气体对外做的功.

题4-9图

解:设KV T =由图可求得直线的斜率K 为

002V T K =

得过程方程

V V T K 00

2=

由状态方程 RT pV υ= 得

V

RT

p υ=

ab 过程气体对外作功

?=0

2d V v V

p A

?

??====00

00

002000220

2

d 2d 2d V V V v V V RT

V V RT V

V V T V R V V RT A

4-10 一卡诺热机在1000 K 和300 K 的两热源之间工作,试计算 (1)热机效率;

(2)若低温热源不变,要使热机效率提高到80%,则高温热源温度需提高多少? (3)若高温热源不变,要使热机效率提高到80%,则低温热源温度需降低多少?

解:(1)卡诺热机效率

121T T -

%701000300

1=-

(2)低温热源温度不变时,若

%80300

11

=-

=T η

要求 15001=T K ,高温热源温度需提高500K (3)高温热源温度不变时,若

%80100012

=-

=T η

要求 2002=T K ,低温热源温度需降低100K

4-11 如题4-11图所示是一理想气体所经历的循环过程,其中AB 和CD 是等压过程,BC 和DA 为绝热过程,已知B 点和C 点的温度分别为2T 和

3T .求此循环效率.这是卡诺循环

? 题4-11图

解: (1)热机效率

12

1Q Q -

=η AB 等压过程 )(12P 1

T T C Q -='υ 吸热

)(P mo 1A B l

T T C M M

Q -=

CD 等压过程 )(12P 2

T T vC Q -='

放热

)(P mol

2

2D C T T C M M

Q Q -='-=

)/1()/1(12B A B C D C A B D C T T T T T T T T T T Q Q --=

--=

根据绝热过程方程得到

AD 绝热过程 γ

γγγ----=D D A A T p T p 11

BC 绝热过程 γ

γγγ----=C C B B T p T p 111

B C D D C B A T T

T T p p p p =

==

23

1T T -=η (2)不是卡诺循环,因为不是工作在两个恒定的热源之间.

4-12 (1)用一卡诺循环的致冷机从7℃的热源中提取1000 J 的热量传向27℃的热源,需要多少功?从-173℃向27℃呢?

(2)一可逆的卡诺机,作热机使用时,如果工作的两热源的温度差愈大,则对于作功就愈有利.当作致冷机使用时,如果两热源的温度差愈大,对于致冷是否也愈有利?为什么? 解:(1)卡诺循环的致冷机

212

2T T T A Q e -=

=

7℃→27℃时,需作功

4.711000280280

30022211=?-=-=

Q T T T A J

173-℃→27℃时,需作功 20001000100100

30022212=?-=-=

Q T T T A J

(2)从上面计算可看到,当高温热源温度一定时,低温热源温度越低,温度差愈大,提取同

样的热量,则所需作功也越多,对致冷是不利的.

4-13 如题4-13图所示,1 mol 双原子分子理想气体,从初态K 300,L 2011==T V 经历三种不同的过程到达末态K 300,L 4022==T V . 图中1→2为等温线,1→4为绝热线,4→2为等压线,1→3为等压线,3→2为等体线.试分别沿这三种过程计算气体的熵变.

题4-13图

解:21→熵变

等温过程 A Q d d = , V p A d d =

RT pV =

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