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Find the total number of collisions between molecules in 1.00 s in 1.00 L of nitrogen gas at standard temperature and pressure ( 0 °C , 1.00 atm). Use 1.88 × 10 −10 m as the effective radius of a nitrogen molecule. (The number of collisions per second is the reciprocal of the collision time.) Keep in mind that each collision involves two molecules, so if one molecule collides once in a certain period of time, the collision of the molecule it hit cannot be counted.

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(a) Estimate the specific heat capacity of sodium from the Law of Dulong and Petit. The molar mass of sodium is 23.0 g/mol. (b) What is the percent error of your estimate from the known value, 1230 J/kg · ° C ?

a. 1080 J/kg °C ; b. 12 %

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A sealed, perfectly insulated container contains 0.630 mol of air at 20.0 °C and an iron stirring bar of mass 40.0 g. The stirring bar is magnetically driven to a kinetic energy of 50.0 J and allowed to slow down by air resistance. What is the equilibrium temperature?

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Find the ratio f ( v p ) / f ( v rms ) for hydrogen gas ( M = 2.02 g/mol ) at a temperature of 77.0 K.

2 e / 3 or about 1.10

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Unreasonable results . (a) Find the temperature of 0.360 kg of water, modeled as an ideal gas, at a pressure of 1.01 × 10 5 Pa if it has a volume of 0.615 m 3 . (b) What is unreasonable about this answer? How could you get a better answer?

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Unreasonable results . (a) Find the average speed of hydrogen sulfide, H 2 S , molecules at a temperature of 250 K. Its molar mass is 31.4 g/mol (b) The result isn’t very unreasonable, but why is it less reliable than those for, say, neon or nitrogen?

a. 411 m/s; b. According to [link] , the C V of H 2 S is significantly different from the theoretical value, so the ideal gas model does not describe it very well at room temperature and pressure, and the Maxwell-Boltzmann speed distribution for ideal gases may not hold very well, even less well at a lower temperature.

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Challenge problems

An airtight dispenser for drinking water is 25 cm × 10 cm in horizontal dimensions and 20 cm tall. It has a tap of negligible volume that opens at the level of the bottom of the dispenser. Initially, it contains water to a level 3.0 cm from the top and air at the ambient pressure, 1.00 atm, from there to the top. When the tap is opened, water will flow out until the gauge pressure at the bottom of the dispenser, and thus at the opening of the tap, is 0. What volume of water flows out? Assume the temperature is constant, the dispenser is perfectly rigid, and the water has a constant density of 1000 kg/m 3 .

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Eight bumper cars, each with a mass of 322 kg, are running in a room 21.0 m long and 13.0 m wide. They have no drivers, so they just bounce around on their own. The rms speed of the cars is 2.50 m/s. Repeating the arguments of Pressure, Temperature, and RMS Speed , find the average force per unit length (analogous to pressure) that the cars exert on the walls.

29.5 N/m

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Verify that v p = 2 k B T m .

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Verify the normalization equation 0 f ( v ) d v = 1 . In doing the integral, first make the substitution u = m 2 k B T v = v v p . This “scaling” transformation gives you all features of the answer except for the integral, which is a dimensionless numerical factor. You’ll need the formula

0 x 2 e x 2 d x = π 4

to find the numerical factor and verify the normalization.

Substituting v = 2 k B T m u and d v = 2 k B T m d u gives
0 4 π ( m 2 k B T ) 3 / 2 v 2 e m v 2 / 2 k B T d v = 0 4 π ( m 2 k B T ) 3 / 2 ( 2 k B T m ) u 2 e u 2 2 k B T m d u = 0 4 π u 2 e u 2 d u = 4 π π 4 = 1

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Verify that v ¯ = 8 π k B T m . Make the same scaling transformation as in the preceding problem.

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Verify that v rms = v 2 = 3 k B T m .

Making the scaling transformation as in the previous problems, we find that
v 2 = 0 4 π ( m 2 k B T ) 3 / 2 v 2 v 2 e m v 2 / 2 k B T d v = 0 4 π 2 k B T m u 4 e u 2 d u .
As in the previous problem, we integrate by parts:
0 u 4 e u 2 d u = [ 1 2 u 3 e u 2 ] 0 + 3 2 0 u 2 e u 2 d u .
Again, the first term is 0, and we were given in an earlier problem that the integral in the second term equals π 4 . We now have
v 2 = 4 π 2 k B T m 3 2 π 4 = 3 k B T m .
Taking the square root of both sides gives the desired result: v rms = 3 k B T m .

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Questions & Answers

What is differential form of Gauss's law?
Rohit Reply
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
OLUWA Reply
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
Abu Reply
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
Lord Reply
kya ye b.sc ka hai agar haa to konsa part
MPL Reply
what is charge quantization
Mayowa Reply
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
Aditya
determine absolute zero
OFERE Reply
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
Opeyemi Reply
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
Ashok Reply
what is the weight of the earth in space
peterpaul Reply
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
Ifeoma Reply
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
Mehmet
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
Mayowa
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
Mehmet
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
Mateshwar Reply
explain the lack of symmetry in the field of the parallel capacitor
Phoebe Reply
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe
Practice Key Terms 3

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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