**Unformatted text preview: **HW #1: Problem 1
Wanted: Determine the ratios of (a) the mean speeds and (b) the mean kinetic energies of H2 and O2 at 25°C.
Given: = 25℃ = 298.15 2 = 2 Solution: < 2 >
=
< 2 >
b) 8
2
8
28 = 1 8
2 1 8
28 2 = 32 = 32 4
=
2
1 1
2 ∗ < 2 >2
< 2 >
2 4
= 2
=
1
< 2 >
32 1
2
2 28 ∗ < 2 > = 8.3145 2 1 1.37 < > = ∗ < >2
2 8 1.32 < > =
a) 2 = ∴ 32
=1
32 < 2 >
4
=
< 2 >
1 ∴ < 2 >
=1
< 2 > HW #1: Problem 2
Wanted: The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At 25˚C and assuming that air
consists of N2 molecules with a collision diameter of 395 pm, calculate (i) the mean speed of the molecules, (ii) the mean
free path, (iii) the collision frequency in the gas.
Given: = 1 = 110−9 k= 1.380610−23
2 = 8.3145 2 = 28 = 25℃ = 298.15 = 395 Solution:
i) 1.32 < > = ii) 8
= 2
2 2
∗ 298.15 = 474.8 ∗ 28
∗ 1000 ∴ < > = 474.8 8 ∗ 8.3145 3
∗ 298.15
<> =
=
=
101325 2
22 2 ∗ 39510−12 2 ∗ 110−9 ∗ 760 1.380610−23 ∴ = 44,551 = 44.5
iii) 2 2 ∗ 474.8 < >
2<>
=
=
=
= 0.0151 −1 = 54.23 ℎ −1 44,551 ∴ 2 = 0.0151 −1 = 54.23 ℎ −1
2 474.8 < >
=
=
= 0.0107 −1 = 38.36 ℎ −1 44,551 HW #1: Problem 3
Wanted: Calculate the percentage of N-2 molecules having speeds between the speed of sound (340.29 m/s) and the
speed of light (2.998*10^8 m/s) at T = 300 K and 1000 K.
Given: 1 = 340.29
2 = 2.998108 Solution: = P = 4 2 1 ∞
0 3
2 ∴ P = 4 P = 4 P = 4 P = 4 2 = 2 F v = 4 1 1 2 − 2 න 2 − 2 ℎ න = 3
2 2 2 − ℎ = ∗ erf 1
3
2 ∗ erf 3
2 3
2 ∗ erf 43/2 2 2 1 −1 erfc 1
+
1 2 2 1 −1
P=
+ erfc T (K) 1 A/π
A(T)
x
300
5.6E-06
1.78E-06
1000 1.68E-06
5.35E-07 T (K) P
300
1000 73.0%
94.3% 2 1 − 2 1 −1
43/2 2 2 2 erfc() 1 − 2 1 −1
43/2 ∗ erfc 0 − − 2 −
43/2 2 − 2 − 2
1
43/2 ∗ erf ∞ − 2 −∞
−
43/2 3
2 2 erfc(x)
0.8053
0.2548
0.4411
0.5329 1 1000 6.0221023
= 2
2 ∗ 1.3810−23 2
∗ 0.00168 2 = 2
28 ∗ ℎ
HW #1: Problem 4
Wanted: What, according to the Maxwell-Boltzmann distribution, is the proportion of gas molecules having (a) more
than, (b) less than the root mean square speed? (c) what are the proportions having speed greater and smaller than the
mean speed?
3
Given:
3
8 2 2 − 2 =
∗ < >=
F v = 4 ℎ = 2
Solution: P = 4 ∴ P = 4 P = 4 P = 4 a)
b) 3
2
3
2 3
2 2 2 − 2 න erf 2 − 2 ℎ න ∗ erf 43/2 erf s = +
3
2 3 3
2 2 −2
3
+
2 8
=
2 P = erf 2 2 ∗ erf − 2 −
43/2 − 2 −
43/2 − 2 − 0
43/2 4 4 4
2 −
4
+ 2 − 2 − A 2 = ∗ erf 0 − 2 0 −0
43/2 2 = erf + 2 − 2 3
2
∗ ℎ
= 0.6083 ∗ ℎ ℎ ℎ ℎ P = 1 − 0.6083 = 0.3917
P = 0.6083 = 0.61 c) = ∗ erf 1 3
=
2 s = P= 3
2 P less than rms = 39%
P more than rms = 61%
A 2 = 4 = 0.5331 P less than the mean = 53%
P more than the mean = 47% P less than rms = 61%
P more than rms = 39% HW #1: Problem 1.1
Wanted: Molecules all of mass m and speed v exert a pressure p on the walls of a vessel. If half the molecules are
replaced by ones of another type all with mass 1/2 m and speed 2v, will the pressure (a) increase, (b) decrease, (c)
remain constant?
Given: P = < >2 ∗
Solution: ( 1) = 1 = 1 < 1 >2 ∗
() =< 1 >2 ∗ + () = 1 + 2 = 1 < 1 >2 ∗ + 1
< 21 >2 ∗ = 31
2 1
< 21 >2 ∗
2 a) Increase in pressure HW #1: Problem 1.3
Wanted: Suppose some property q of a gas is proportional to (0.326 (s^3/m^3)Vx^3+((pi)(s^9/m^9)Vx^9. What is the
average value of q?
Given:
∞
< > = න −∞ Solution: ∞ < > = න = 0.3263 + 9 =
−∞ 0.3264 10 ∞
+
=∞−∞=0
4
10 −∞ < >=0 HW #1: Problem 1.8
Wanted: By setting the derivative of the formula for the Maxwell-Boltzmann speed distribution equal to zero, show that
the speed at which the distribution has its maximum is given by equation 1.33.
3
Given: 2 2 − 2 F v = 4 ℎ = 2
Solution: F v = 4 3
2 2 − 2 − 2
2 2 = 2 + − 2 = −2 − ( 2 − 1) = 0 2 ∴ 0 = −2 − ( 2 − 1) ∴ = 1
= 2 2 2 3 2 − = 2 2 − 2 2 = 1 HW #1: Problem 1.9
Wanted: Show using equations 1.16 and 1.37 that the average molecular energy is 3kT/2.
Given:
∞ 1.16 < > = න = 1 1.37 = 2
0 Solution: = 2 < > = 2 ∴ < > = 2 3/2 3/2 − 0 න 3/2 − ℎ න 3/2 − = − < > = න ∞ 0
3/2 3/2 ∞ 1 ℎ = ∞ 1 0 3 4 5/2 3 6 ()
3
=
=
3/2
4 4 ()
2 ∴< >= 3
2 HW #1: Problem 1.15
Wanted: Calculate the root-mean-squared deviation of the speed from its mean value: [<(v - <v>)^2>]^1/2.
Given:
[(− < >)2 ] 1/2 = [< 2 > − < >2 ]1/2
8
3
< >=
< 2 > = Solution: 3 8
− [< 2 > − < >2 ]1/2 = 1/2 = 1/2 8
3− = 0.454 ∗ ∞ ∴ [< 2 > − < >2 ]1/2 = 0.454 ∗ [(− < >)2 ] = න [(− < >)]
0 ∞ = න 4 3
0 [(− < >)2 ] 1/2 = [< 2 > − < >2 ]1/2 HW #1: Problem 1.10 (BONUS)
Wanted: Prove equation 1.41 from equation 1.40.
Given:
−
1 3/2 ∞
∗ 1.40 = 2
න ∗ = 2 , ∗ = 2 , = 2 3/2 1 3/2 ∗ = 2 − 2 − 2 (− < >)
2 1 3/2 0 − = , ∞ න 1.41 ∗ =
=
∞ − ℎ න ∗ 1
, − = ∗ erfc
2 − − = 2 1 3/2 2
2 − + erfc() = 2 erfc
2 erfc
2 − − + 2 − ∗ ∗ = 2 ∗ − + erfc 2 ∴ ∗ 2 −
=
+ erfc ∗ = ∗ * Show more steps for full credit ...

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- Spring '18