EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9781305804463
Author: Jewett
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 21, Problem 21.61AP

(a)

To determine

The rms speed for a particle of diameter d .

(a)

Expert Solution
Check Mark

Answer to Problem 21.61AP

The rms speed for a particle of diameter d is (4.82×1012)d32m/s .

Explanation of Solution

Given info: Density of spherical particle is 1.00×103kg/m3 , temperature is 20.0°C and diameter of particle is d .

Write the expression for the rms speed:

vrms=3kBTm (1)

Here,

vrms is rms speed of particle.

kB is boltzmann’s constant.

T is temperature.

m is mass of gas molecule.

Write the formula for mass:

m=ρ×V (2)

Here,

ρ is density of spherical particle.

V is volume of spherical particle.

Write the formula for volume of spherical particle:

V=43πr3 (3)

It is given that diameter of spherical particle is d . So, the radius of spherical particle is d2 .

Substitute 3.14 for π and d2m for r in equation (3).

V=43(3.14)(d2m)3=0.52d3m3

Substitute 1.00×103kg/m3 for ρ and 0.52d3m3 for V in equation (2).

m=(1.00×103kg/m3)(0.52d3m3)=520d3kg

Since, boltzmann’s constant is 1.38×1023J/K .

Substitute 520d3kg for m , 20.0°C for T and 1.38×1023J/K for kB in equation (1).

vrms=3(1.38×1023J/K×1kgm2/s21J)(20.0+273K)520d3kg=(4.82×1012)d32m/s

Conclusion:

Therefore, the rms speed for a particle of diameter d is (4.82×1012)d32m/s .

(b)

To determine

The time interval for particle to move a distance equal to its own diameter.

(b)

Expert Solution
Check Mark

Answer to Problem 21.61AP

The time interval for particle to move a distance equal to its own diameter is (2.08×1011)d52s .

Explanation of Solution

Given info: Density of spherical particle is 1.00×103kg/m3 and temperature is 20.0°C .

Write the expression for the time interval related to rms:

t=dvrms (4)

Here,

t is time interval.

d is distance.

vrms is rms speed.

Since particle is moving equal to its diameter. So, the distance travelled by the particles is d .

Substitute d for d and (4.82×1012)d32m/s for vrms as calculated in above part in equation (3).

t=dm(4.82×1012)d32m/s=(2.08×1011)d52s

Conclusion:

Therefore, the time interval for particle to move a distance equal to its own diameter is (2.08×1011)d52s .

(c)

To determine

The rms speed and the time interval for a particle of diameter 3.00μm .

(c)

Expert Solution
Check Mark

Answer to Problem 21.61AP

The rms speed for a particle of diameter 3.00μm is 0.926mm/s and the time interval for a particle of diameter 3.00μm is 3.24ms .

Explanation of Solution

Given info: Density of spherical particle is 1.00×103kg/m3 , temperature is 20.0°C and diameter of particle is 3.00μm .

The rms speed for a particle of diameter d as calculated in above part is,

vrms= (4.82×1012)d32m/s

Substitute 3.00μm for d in above equation.

vrms=(4.82×1012)(3.00μm×106m1μm)32m/s=9.27×104m/s×103mm1m0.926mm/s

The time interval for a particle of diameter d as calculated in above part is,

t=(2.08×1011)d52s

Substitute 3.00μm for d in above equation.

t=(2.08×1011)(3.00μm×106m1μm)52m1s=0.00324s×103ms1s=3.24ms

Conclusion:

Therefore, the rms speed for a particle of diameter 3.00μm is 0.926mm/s and the time interval for a particle of diameter 3.00μm is 3.24ms .

(d)

To determine

The rms speed and the time interval for a sphere of 70.0kg .

(d)

Expert Solution
Check Mark

Answer to Problem 21.61AP

The rms speed for a sphere of 70.0kg is 1.32×1011m/s and the time interval for a sphere of 70.0kg is 3.88×1010s .

Explanation of Solution

Given info: Density of spherical particle is 1.00×103kg/m3 , temperature is 20.0°C and mass of the sphere is 70.0kg .

Write the expression for the rms speed:

vrms=3kBTm (5)

Here,

vrms is rms speed of particle.

kB is boltzmann’s constant.

T is temperature.

m is mass of sphere.

Substitute 70.0kg for m , 20.0°C for T and 1.38×1023J/K for kB in equation (5).

vrms=3(1.38×1023J/K×1kgm2/s21J)(20.0+273K)70.0kg=1.32×1011m/s

Thus, the rms speed of the partical is 1.32×1011m/s .

Write the formaula for volume in relation to mass and density.

V=mD

Here,

D is density.

Substitute 70.0kg for m and 1.00×103kg/m3 in above equation.

V=70.0kg1.00×103kg/m3=0.07m3

Write the formula for the volume of sphere:

V=43π(d2)3

Here,

V is volume of sphere.

d is diameter of sphere.

Rearrange above equation for d .

(d2)3=3V4πd=2(3V4π)13 (6)

Substitute 0.07m3 for V and 3.14 for π in equation (6).

d=2(3×0.07m34×3.14)13=0.51m

Write the expression for the time interval related to rms:

t=dvrms

Substitute d for d in above equation.

t=dvrms

Substitute 0.51m for d and 1.32×1011m/s for vrms in above equation.

t=0.51m1.32×1011m/s=3.86×1010s

Conclusion:

Therefore, the rms speed for a sphere of 70.0kg is 1.32×1011m/s and the time interval for a sphere of 70.0kg is 3.88×1010s .

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Chapter 21 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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