2010-2011学年春季《热学》习题3（答案）

2010-2011学年春季《热学》习题3（答案）
第二章分子动理学理论的平衡态理论
习题三：
1. 质量为10Kg 的氮气，当压强为1.0atm,体积为7700cm 3 时，其分子的平均平动能是多少？
解: ∵mR
PV T m
M =
而 T 2
3k =
ε
∴
J
10
4.5)
10022.6()10(2)
028.0()107700()10013.1(3232324
1
23
1
36
5
----=
=
=
mol
Kg mol
Kg m Pa mN
PVM mR
kPVM A
m m
ε
2. 求常温下质量为m 1=
3.00g 的水蒸气与m 2=3.00g 的氢气组成的混合理想气体的定体摩尔热容. 设：水蒸汽的定容比热为1V c ，氢气的定容比热为2V c ，则混合气体的定容比热可由量热学 ,21Q Q Q += 即 221121)(M c M c c M M V V V +=+，得
)1()
(212
211 M M M c M c c V V V ++=
，定容摩尔热容量)2( V mol V c M C =，
对混合气体，其平均摩尔质量满足关系：
)3(1112
2
11
M
M M
M
M M
M
mol mol mol
+
=
（注：（3）式可由理想气体状态方程及道尔顿分压定律得出。

）综合上述三式，其中21M M M +=，可得
M
M M M M M M c M c M C mol mol V V V
221122111
11
+=
）（，令：水蒸汽和氢气的定容摩尔热容量分别为111 mol V V M c C =、222mol V V M c C =，则上式变为22
11
22
2111
m o l
m o l
m o l
V m o l
V V M
M M
M M
M C M M C C +
+=
，
对水蒸汽，;32
61R k N C A V =?
=对氢气，R k N C A V 2
52
52=
=，故有
)K J/mol (21.2120
512
318
323
251833?==+
+?=
R R R C V
3. 在等温大气模式中，设气温为5o
C ，同时测得海平面的大气压和山顶的气压分别为a P 1000.15?和
a P 1078.05
，试问山顶的海拔为多少？
解：根据玻尔兹曼分布律，可得等温高度公式，代入数据，得
z =RT M m g ln p 0p =(8.31J ×mol -1×K -1)′(278K )(29′10-3Kg ×mol -1)′(9.8m ×s -2)ln 1.00′105Pa 0.78′105
Pa
=2019.7(m)。

4. 当液体与其饱和蒸气共存时，气化率与凝结率相等。

设所有碰到液面上的蒸气分子都能凝结为液
体，并假定当把液面上的蒸气迅速抽去时，液体的气化率与存在饱和蒸气时的气化率相同。

已知水银在0o C 时的饱和蒸气压为2N 0246.0-?m ，问每秒通过每平方厘米液面有多少克水银向真空中气化。

解：以饱和蒸汽为研究对象，设所有碰到液面的蒸汽分子都凝结为液体，则t ?内碰到A ?液面的气体质量为：
t A M RT
kT
p N M t A v n N M t A m M m
A
m A
m =
=
Γ?=π8414
1 代入数据：
M m =200.5′10-3kg ×mol -1,p =0.0246p a ,D A =10-4m 2,T
=273K M =(200.5′10-3kg ×mol -1)′(0.0246Pa )′(10-4m 2)4′(8.31J ×mol -1×K -1)′(273K )′4
′
8′(8.31J ×mol -1×K -1)′(273K )3.1416′(200.5Kg ×mol -1)
得 )(10
23.99
kg M -?=。

按题意，假设凝结率等于汽化率，则该结果也就是每秒通过每平方厘米液面逸出的水银分子的质量。

5. 某种气体的分子由四个原子组成，它们分别处在正四面体的四个顶点上。

（1）求这种分子的平动、
转动和振动自由度数；（2）根据能量均分定理求这种气体的定容摩尔热容。

解：
（1）因n 个原子组成的分子最多有3n 个自由度。

其中3个平动自由度，3个转动自由度，3n-6个是振动自由度，这里n=4，12433=?==n i ，故有12个自由度。

（2）作刚性近似，自由度数6=i ，)K J/mol (93.243?==R C V
6. A sealed cubical container 20.0 cm on a side contains three times Avogadro ’s number of molecules
at a temperature of 20.0℃. Find the force exerted by the gas on one of the walls of the container . [Answer] Now 1 2310023.6-?=mol N A , 1
1
31.8--??=K
mol
J R , K T 15.293=
the mole of the gas mol N
N N N n A A
33===
, the side length of the container m L 2.0=
One-sided area of the container 22204.0)2.0(m m L A === The volume of the container 333008.0)2.0(m m L V === Assuming the gas is ideal, according to the Equation of State for an ideal gas, nRT pV =
So p =nRT V =(3mol )×(8.31J ×mol -1×K -1)×(293.15K )
0.008m
3=9.135′105J ×m -3=9.135′105Pa Then, the force exerted by the gas on one of the walls of the container
N m Pa m Pa pA F 4
2
2
5
10654.33654004.0)10135.9(?=?=??==
7. In a period of 1.00 s, 5.00×1023 nitrogen molecules strike
a wall with an area of 8.00 cm 2
. If the
molecules move with a speed of 300 m/s and strike the wall head on in perfectly elastic collisions, what is the pressure exerted on the wall? (The mass of one N 2 mol ecule is 4.68×10-26 kg.) [Answer] the mole of nitrogen molecules mol mol
n 830.010
022.61000.51
23
23
=??=-
The total mass of nitrogen molecules
Kg Kg Nm m sum 0234.0)10
68.4()10
00.5(26
23
Assuming the velocity attitude of nitrogen molecules before collision is the positive direction, so the speeds of nitrogen molecules before collision and after collision are s m v /300= and s m v /300'
-=, respectively.
According to the Momentum theorem, the impulse of single nitrogen molecule caused by the wall is
1
23
26
'
/
10
808.2)/300/300()10
68.4()(----=--??=-=-=s
m Kg s m s m Kg v v m mv mv p
The minus sign represents the direction of the impulse of nitrogen molecule is opposite to that of the velocity.
The total impulse of all nitrogen molecules is
1
123
23
04.14)10
808.2()10
00.5(---??-=-??==s
m Kg s m Kg Np p sum
The force of the nitrogen molecules caused by wall is
N s
m Kg s s m Kg t
N 04.1404.14)
00.1()
04.14(2
1
-=??-=??-=
=
--
According to The Newton third law, the force exerted by the nitrogen molecules on the wall is
N F F N w 04.14=-=
Since the area of the wall strike by nitrogen molecules 241000.8m A -?= The pressure exerted on the wall
Pa m
N m
N A F P w 4
2
42
4
10755.110755.110
00.804.14?=??=?=
=
--
8. In a constant-volume process, 209J of energy is transferred by heat to 1.00mol of an ideal
monatomic gas initially at 300K. Find (1) the increase in internal energy of the gas. (2) The work it does. (3) Its final temperature.
[Answer] (1) In the constant-volume process, the energy was transformed into internal energy, so the increase of the internal
energy is 209J.
(2) In the constant-volume process, the system does not do work, so the work it does is 0J. (3) 1
1
31.8,2092
3U --??==?=?K
mol
J R J T R m
So, K K
mol
J mol 77.16)
31.8(2
31J 209T 1
1
==
-- K 77.316K 77.16K 300T =+= So the final temperature is 316.77K
9. One mole of an ideal diatomic gas with 2/5R C v = occupies a volume i V at a pressure i P . The
gas undergoes a process in which the pressure is proportional to the volume. At the end of the process, it is found that rms v of the gas molecules has doubled from its initial value. Determine the amount of energy transferred to the gas by heat. [Answer] T T T T M RT v m
rms 343=-=??=
1,2
152
5E ==?=
=n nRT T Rn T n C v trans
According to the equation of state for an ideal gas, nRT V i
=i P So i i trans V P 215E =
10. The dimensions of a room are 4.20m ×3.00m ×2.50m. (1) Find the number of molecules of air in it
at atmospheric pressure and 20.0℃. (2) Find the mass of this air, assuming that the air consists of diatomic molecules with a molar mass of 28.9g/mol. (3) Find the average kinetic energy of a molecule. (4) Find the root-mean-square molecular speed. (5) On the assumption that the specific heat is a constant independent of temperature, we have 2/5int nRT E =. Find the internal energy in the air. (6) Find the internal energy of the air in the room at 25.0℃. [Answer]
(1) P =1.013′105P a ,R =8.31J ×mol -1×K -1,T =293.15K 350.3150.200.320.4V m m m m =??=
According to the equation of state for an ideal gas, nRT =PV m o l K K
m o l J m Pa RT
PV n 87.1309)
15.293)(31.8()50.31)(10013.1(1
1
3
5
=
=
--
a t o m s
m o l a t o m s m o l nN A
26
123
10
88.7)10022.6()87.1309(N ?===- (2) g mol g mol nM
m m
4
1
1078.3)9.28()87.1309(?≈??==- (3) 1
1
1
1
2
76.214)
9.28(1416.3)
15.293()31.8(88----?≈=
=
g
J mol
g K K
mol
J M RT
v m
π
So, J g J mol g v N M v m A m k 21 23
1
1
2
2
10
15.510
022.62)
76.214)(9.28(2121
E ---?≈=?==
(4) s m mol
g K K
mol
J M RT v m
rms /02.5029.28)
15.293()31.8(331
1
1
≈=
=
---
(5) J K K
mol
J mol nRT 6
1
1
int 1098.72
)
15.293()31.8()87.1309(52
5E ?≈=
=
--
(6) According to the equation of state for an ideal gas, nRT =PV
)tan (V t cons P
nRT =
So,
2
21
1P T P T =
, Pa K
K Pa 5
5
1
2121003.115.293)
15.298)(10013.1(T T P P ?≈?=
=
J m Pa V P nRT 6
3
5
221024.3)50.31()1003.1(?≈??==? J J nRT 6
6
2
int 1010.82
)
1024.3(52
5E ?=??=
=。