Ic_class_100201

course Phy: 202

Ic_class_100201Things you should know:

P V = n R T

1/2 rho v^2 + rho g y + P is constant

KE_ave = 3/2 k T

A gram of water requires about 4 Joules of energy to raise its temperature by 1 Celsius degree.

A Fahrenheit degree is 5/9 of a Celsius degree, a Kelvin degree and a Celsius degree are equal, water freezes at 32 F or 0 Celsius or 273 Kelvin.

The temperature of an object is uniform when the average thermal energy per particle, throughout all parts of the object, is uniform. The average thermal energy per particle is 3/2 k T and does not depend on the mass of the particle.

Pa is the unit of pressure. Pa is shorthand for N / m^2.

A liter is not an SI unit of measure. A liter is .001 m^3. The m^3 is the SI unit of measure. When using P V = n R T, if you use volume in liters you will have some potentially confusing conversions to deal with. If you replace the unit 'liter' with '.001 m^3' then your units will work out more easily. The same comment applies to cm^3; a cm^3 is .001 liter, which is .001 ( .001 m^3) = .000001 m^3, or 10^-6 m^3.

The density of water is 1000 kg / m^3, or 1 kg / liter, or 1 gram / cm^3. All three units are equal: Since a m^3 is 1000 liters, 1000 kg / m^3 = 1000 kg / (1000 liters) = 1 kg / liter. Since a liter is 1000 cm^3, 1 kg / liter = 1 kg / (1000 cm^3) = .001 kg / cm^3 = 1 gram / cm^3. We can also say that since a m^3 is 1 000 000 cm^3, 1000 kg / m^3 is 1000 kg / (1 000 000 cm^3) = .001 kg / cm^3 = 1 gram / cm^3.

Rules of thumb:

A 10 cm column of water is supported by a pressure of about 1000 Pa.

Atmospheric pressure is about 100 000 Pa.

Water in a tube

If you hold a tube with two open ends, partially filled with water, in such a way that the ends are at the same level, you will see that the water level on both sides will always 'seek' an equilibrium position in which the two levels are the same. If you quickly raise one end (let's say the end you raise is the right end) the 'middle' of the tube will rise further from the floor and, even though the left end of the tube will stay where it is, the water level on both sides will increase relative to the floor.

• Because friction between water and tube will exert a net upward force on the water on the raised side, the water on that side will be accelerated more than the water on the stationary side, so the water level on the raised side will at first increase more that the water level on the other. Thus the water on the raised side will for at least a short time be higher than that on the stationary side.

• The force exerted by gravity on the water in the 'higher' side will therefore exceed the force exerted by gravity on water in the 'lower' side. That is, gravity will be pulling one side down with more force than the other. If the tube remains open at both ends, the force exerted by air pressure is the same on both sides. The result is a greater net force on one side than on the other, and the water in the tube will therefore be accelerated toward the 'lower' side.

• This will continue until the water level is the same on both sides, at which point the water (which has been accelerating) will be moving at a nonzero velocity toward the lower side. The water will therefore continue moving past the equilibrium point (i.e., the point at which the water levels are equal), and the level on the left side will therefore continue to increase until frictional forces between water and tube have slowed it to rest, at which point the water will begin accelerating in the opposite direction.

• The water level will therefore oscillate about the equilibrium level, but with decreasing amplitude (due to dissipation of energy by friction), until it reaches a steady state in which the levels on both sides are again equal.

`q001. You observed this when the tube was passed around the room. Describe what you observed, and detail whether your observation was consistent with the above description of the process.

****

When one end was raised a little higher than the other, the water column would shift and the two ends of the water would be equal. This is consistant with the description above; as the system is moved or shifted, the water column shifts in the correct direction to “seek an equilibrium point.

#$&*

`q002. Assume the tube contains 10 cm^3 of water. If its cross-sectional area is .3 cm^2, and if the water on one side is 5 cm higher than on the other, what is the net gravitational force on the water (hint: how much does a 5 cm column of water weigh)? What therefore will be the acceleration of the water in the tube, at this instant?

****

5 cm * 0.3 cm2 = 1.5 cm3 or 1.5 mL

1 gram = 1 mL

So 0.0015 kilograms in 5 cm length

F = ma

F = 0.0015 kilograms * 980m/s2

F = 1.47 kg * m/s2 or 1.47 Newtons

#$&*

`q003. Suppose you start with the water level equal in both sides of the tube, the ends both open. You raise one end a few cm, and wait for the level to again equalize. What happens to the length of the air column on the raised side? Does it get longer or shorter? Explain your thinking.

****

The air column will get shorter because the water column in that end is getting longer, so the air must get shorter.

#$&*

`q004. Now suppose you cap one end of the tube, with water level the same on both sides, and raise that end. The air column in that end is now trapped.

Will the length of the air column increase or decrease?

****

The air column will decrease, because the pressure will build up.

#$&*

Will this cause the pressure in the column to increase or decrease?

****

The pressure will increase.

#$&*

Will this therefore cause air pressure to exert more or less force on the water on the raised side?

****

This will cause more force to be exerted on the water column.

#$&*

Is this consistent with the observed relative level of water in the two sides?

****

This is somewhat consistent but the end isn’t sealed so the pressure won’t change.

#$&*

`q005. Suppose you cap the right end of the tube. You then raise one end in such a way as to shorten the trapped air column by 3%.

Assuming that atmospheric pressure is 100 000 Pa, what is the pressure in the shortened column?

****

100,000 Pa * 3% = 3000 Pa

100,000 Pa – 3000 Pa = 97,000 Pa

#$&*

On which side of the system will the water level be higher? How much higher?

****

The system where it is uncapped. The amount of difference depends on how far you raise one side of the tube. The first system can be raised until the water flows out the end. However on the second system, the pressure can only increase so much so the water column can only increase so much.

#$&*

Show that the net force on the water in the tube is zero.

****

‘dKE + ‘dPE + ‘dW_NC_BY = 0

#$&*

`q006. An air column is trapped between a water column and the end of a tube with uniform, unchanging cross-sectional area.

If the length of the air column becomes 10% greater, then what is the corresponding volume ratio (i.e., the ratio of the volume after to the volume before)?

****

1:10

Length corresponds to pressure. Volume and pressure are constant so that means if the length goes up, pressure must go up, then volume must go up by 10%

#$&*

What is the corresponding pressure ratio (the ratio of pressure after to pressure before)?

****

1:10

Length corresponds to pressure. Pressure and volume are constant so that means if the volume goes up, length goes up, then pressure must go up by 10%

#$&*

Answer the same questions if the length of the air column becomes 30% shorter.

****

10:7

Length corresponds to pressure. Volume and pressure are constant so that means if the length goes up, pressure must go up, then volume must go up by 30%

For instance if the volume was 10 mL and it decreased by 30% then the final volume would be 7 mL.

#$&*

If the original pressure was 100 000 Pa, what are the final pressures in these two situations?

****

1st system final pressure = 90,000 Pa

2nd system final pressure = 70,000 Pa

#$&*

2-bottle system

`q007. Using a 2-bottle system, with the 'long' tube connecting the two bottles full of water with its end under water in both tubes, you observed that when the 'pressure tubes' and 'release tubes' are capped and one bottle is raised the water level in that bottle decreases, while the level in the other bottle increases. You took data to determine the change in volume in the lower bottle.

What were your data for this activity?

****

Trial One Trail Two

Air Column Before 45 cm 45 cm

Air Column After 43 cm 41 cm

Water Column 67 cm 67 cm

#$&*

What do you conclude was the volume change of the gas in the bottle?

****

Trial One

L = 2 cm

CS = .3cm2

Volume = 0.6 cm3 or 0.6 mL

Trial Two

L = 4 cm

CS = .3cm2

Volume = 1.2 cm3 or 1.2 mL

#$&*

What is the volume change as a percent of the original volume?

****

Original

L = 45 cm

CS = .3cm2

Volume = 13.5 cm3 or 13.5 mL

Trial One

0.6 mL / 13.5 mL = 0.044 %

Trial Two

1.2 mL / 13.5 mL = 0.089 %

#$&*

What do you conclude is the pressure change in the bottle?

****

The pressure change is insignificant to the original pressure.

#$&*

What was the height of the water column in the vertical tube connecting the two bottles? What is the vertical distance between the water surfaces in the two bottles?

****

45.5 cm, 15.25 cm

#$&*

What therefore must be the pressure difference between the water surface in the lower bottle, and the water surface in the upper bottle?

****

1525 Pa

There's about a 30 cm difference in altitude, which should give you about a 3000 Pa difference in pressure.

#$&*

Is this difference accounted for by the pressure change you calculated based on the change in the water level in the lower bottle?

****

Yes the pressure change is calculated by the change in the lower bottle.

#$&*

How would you expect the air columns in the pressure tubes to behave in this experiment?

****

They would shorten as the pressure increased and get longer as the pressure decreased.

#$&*

Would the air columns both get shorter, both longer, or would one get shorter and the other longer? Explain.

****

One would get longer and one would get shorter because as you raise a bottle, the volume decreases, so the pressure decreases. The other bottle will remain still making it lower than the first and as the volume will increase, so the pressure will increase.

#$&*

`q008. You did the same activity with the second bottle 'uncapped'. What differences did you observe and how do you interpret them?

****

The water levels changed quite a bit, but the air columns didn’t change that much because the pressure wasn’t in a closed space; it could escape.

#$&*

Beads dropping on pan balance

`q009. Beads of mass m_bead are dropped onto the pan of the balance hitting the balance at velocity v and rebounding at velocity v '. The average time between drops is `dt. Using symbols, reason out the momentum change experienced by a single bead, and the average force exerted by the dropped beads.

What is the symbolic expression for the momentum change?

****

P = mv

#$&*

What is the symbolic expression for the average force?

****

F_ave = ma

#$&*

`q010. Suppose the velocities v and v ' of the bead in the preceding are respectively -4 m/s and 2 m/s, and that `dt is .8 second. Substitute these values into the symbolic expression you obtained in the preceding. What is your result (note that m_bead is not given, so m_bead will remain in its symbolic form)?

****

P = mv

P = m_bead * -1 m/s

you might be confusing the symbol for momentum with the symbol for pressure

F_ave = ma

F_ave = m_bead * -2.5 m/s2

#$&*

What is the weight of a bead whose mass is m_bead?

****

m_bead * 9.8 cm/s2

#$&*

Compare your two expressions. How many times the weight of a bead is the average force?

****

m_bead * 980 m/s2 = m_bead * -2.5 m/s2

980 m/s2 / -2.5 m/s2 = about 400 times, approximately 392 times

#$&*

Energy in a gas

A monatomic gas consists of atoms which behave as particles. Their average KE is 3/2 k T, where k = R / N_A and T is absolute temperature.

• A system which consists of n moles of gas has n * N_A particles, so the total KE of all these particles is n * N_A * (3/2 k T).

• n * N_A * (3/2 k T) in simlified form can be written as 3/2 n * N_A * k * T.

Since k = R / N_A this can be written as 3/2 n * N_A * R / N_A * T, which simplifies to

3/2 n R T.

We conclude that the total kinetic energy of the particles in n moles of a monatomic gas is 3/2 n R T.

We call this the total internal energy of the gas and denote it by the symbol U. We write

• U = 3/2 n R T, for a monatomic gas.

For a reason related to the frequent behavior of a dropped pony bead, which usually rebounds to about 70% of its original height but which occasionally rebounds to only a few percent of the original height before falling back to the floor then rebounding much higher than before, a mole of diatomic gas (air, for example, consisting predominantly of diatomic nitrogen and diatomic oxygen and hence behaving pretty much as a diatomic gas) at a given temperature has 5/3 as much internal energy as a mole of monatomic gas at the same temperature. 5/3 * 3/2 n R T = 5/2 n R T, so

• U = 5/2 n R T for a diatomic gas.

(we'll see the connection between this and the behavior of the bead in our next class).

`q011. If the temperature of n moles of monatomic gas changes from T1 to T2, then what is the expression for the change U2 - U1 in internal energy of the gas?

****

T1/T2 = U1/U2

U2 / U1 is not the same as `dU = U2 - U1.

#$&*

What would be the change if the gas was diatomic?

****

T1/T2 = 2(U1/U2)

#$&*

`q012. 2 liters of confined diatomic gas in a bottle is heated from 0 Celsius and atmospheric pressure to 50 Celsius, at constant volume. What is the pressure at 50 Celsius, and what is the height of a water column that could be supported by that pressure?

****

T1/T2 = P2/P1

273 K/323 K =? P2/ 1 Atm

273 K * Atm = 323K*? P2

P2 = 0.84 Atm

The pressure doesn't decrease with increasing temperature. If P / T is constant, then T2 / T1 = P2 / P1, which differs from your T1/T2 = P2/P1. If you correct this reversal you'll find that pressure is 1.18 atm.

1 atm = 101.3 kPa, so

1 atm/101.3 kPa = 0.84 atm/? P2

85.092 kPa*atm = 1 atm*? P2

P2 = 85.092 kPa, so 0.85092 cm of water will be supported by this pressure.

#$&*

The gas is then heated to 100 Celsius, at constant pressure. What will be its volume at this temperature?

****

T1/T2 = U2/U1

323K/423 K =? U2/ 2 Liters

646 K*Liters = 423 K*? U2

U2 = 1.53 Liters

#$&*

The expansion of the gas displaces an amount of water equal to the change in gas volume, to the height you found in the first part of the question. By how much does the PE of the system therefore increase?

****

I thought the potential energy would decrease because the volume and temperature are inversely proportional. I’m not sure but my work is below!

PE = mgh

PE = 2 Liters * 9.8 cm/s2 * 1 cm

PE = 19.6 Joules

PE = mgh

PE = 2 Liters * 9.8 cm/s2 * 0.85092 cm

PE = 16.67 Joules

PE = 19.6 Joules – 16.67 Joules

PE = 2.93 Joules

#$&*

By how much does the internal energy of the gas change as we heat it from 0 C to 100 C?

****

U1 = 3/2 T, because n and R are constant!

U1 = 3/2 (273K)

U1 = 409.5 K

U2 = 3/2 T

U2 = 3/2 (373K)

U2 = 559.5

U2 – U1 = 559.5 K – 409.5 K

‘dU = 150 K

#$&*

What is the PE change of the system as a percent of the internal energy change of the gas?

****

‘dPE = 2.93 Joules

‘dU = 150 K

2.93 joules/ 150 K = 1.95%

#$&*

________________________________________

Homework:

Submit the above questions in the usual manner.

Within the next week you should master the problems in Introductory Problem Set 5, found at http://vhmthphy.vhcc.edu/ph2introsets/default.htm .

Good answers on several questions, but be sure you compare your work with the notes below:

Water in a tube

If you hold a tube with two open ends, partially filled

with water, in such a way that the ends are at the same level, you will see that

the water level on both sides will always 'seek' an equilibrium position in

which the two levels are the same.  If you quickly raise one end (let's say

the end you raise is the right end) the 'middle' of the tube will rise further

from the floor and, even though the left end of the tube will stay where it is,

the water level on both sides will increase relative to the floor. 

`q001.  You observed this when the tube was passed

around the room.  Describe what you observed, and detail whether your

observation was consistent with the above description of the process.

****

#$&*

`q002.  Assume the tube contains 10 cm^3 of water. 

If its cross-sectional area is .3 cm^2, and if the water on one side is 5 cm

higher than on the other, what is the net gravitational force on the water

(hint:  how much does a 5 cm column of water weigh)?  What therefore

will be the acceleration of the water in the tube, at this instant?

****

The mass of 10 cm^3 of water is 10 grams or .01 kg.

The 5 cm column has volume V = A * h = .3 cm^2 + 5 cm =

1.5 cm^3.  Its mass is therefore 1.5 grams, or .0015 kg.

The weight of the column is the force exerted on it by

gravity.  Since gravity accelerates masses at 9.8 m/s^2, we get

The weight of the water on the higher side therefore

exceeds the weight on the lower side by .015 N, and this is the net

gravitational force on the system.

If this is the net force on the .01 kg mass of the column,

it will therefore accelerate at

#$&*

`q003.  Suppose you start with the water level equal

in both sides of the tube, the ends both open.  You raise one end a few cm,

and wait for the level to again equalize.  What happens to the length of

the air column on the raised side?  Does it get longer or shorter? 

Explain your thinking.

****

When you raise one end of the tube, the lowest point on

the tube also goes higher, so the water level relative to the floor will

increase. 

The increase in water level relative to the floor will be

half as great as distance the one side was raised.

The air column on the raised side originally extended from

the first water level to the end of the tube.  When that side was raised,

the end of the tube was raised relative to the floor and the water level

increased by half this distance.  So the air column on that side was raised

half as far as was the end of the tube.

#$&*

`q004.  Now suppose you cap one end of the tube, with

water level the same on both sides, and raise that end.  The air column in

that end is now trapped. 

Will the length of the air column increase or decrease? 

****

You have capped off the end to be raised.  As you

raise this end the water column tends to lengthen, but in doing so it would

increase length of the air column on that side.  This would lead to lower

pressure.  ...

#$&*

Will this cause the pressure in the column to increase or

decrease? 

****

#$&*

Will this therefore cause air pressure to exert more or

less force on the water on the raised side? 

****

#$&*

Is this consistent with the observed relative level of

water in the two sides?

****

#$&*

`q005.  Suppose you cap the right end of the tube. 

You then raise one end in such a way as to shorten the trapped air column by 3%. 

Assuming that atmospheric pressure is 100 000 Pa, what is

the pressure in the shortened column? 

****

If the air column is shortened by 3%, its volume decreases

by 3%, to .97 times its original volume.

This increases the pressure.  Assuming that T and n

remain constant, we get PV = constant so that P0 * V0 = P1 * V1.

It follows that

P1 = P0 * (V0 / V1) = P0 * (V0 / (.97 V0) ) = P0 * 1.031,

approx..

The 3% decrease in volume results in very nearly a 3%

increase in pressure (note that the increase is not exactly 3%, since 1 / .97 is

a little more than 1.03).

If the pressure raises by 3%, then it change by a factor

of 1.03, to 103 000 Pa.

#$&*

On which side of the system will the water level be

higher?  How much higher? 

****

To shorten the air column on the capped end you would need

to raise the opposite end.  The water level would increase on both sides,

but due to the increase in the pressure of the air column the pressure exerted

on the water column at the raised end will exceed that on the open end.  To

maintain equilibrium the water column in the open end will therefore need to be

higher than on the closed end.

#$&*

Show that the net force on the water in the tube is zero.

****

It must be so, once the water comes to equilibrium.

#$&*

`q006.  An air column is trapped between a water

column and the end of a tube with uniform, unchanging cross-sectional area. 

If the length of the air column becomes 10% greater, then

what is the corresponding volume ratio (i.e., the ratio of the volume after to

the volume before)? 

****

The cross-sectional area is unchanging, so with a 10%

increase in column length the volume will become 10% greater.  The volume

ratio will therefore be 1.10.

#$&*

What is the corresponding pressure ratio (the ratio of

pressure after to pressure before)? 

****

Since n and T remain constant, P V remains constant, so P0

V0 = P1 V1 and we have

P1 = P0 * (V0 / V1) = P0 * ( V0 / (1.10 V0) ) = .91 P0.

The 10% increase in volume results in a 9% decrease in

pressure.

#$&*

Answer the same questions if the length of the air column

becomes 30% shorter. 

****

Reasoning similar to the above leads us to the conclusion

that the volume decreases by 30%, so 0.70 of its original value, so that

P1 = P0 * (V0 / (0.70 V0) ) = 1.43 P0, approx..

The 30% decrease in length results in an increase of about

43% in pressure.

Based on the rules of thumb we used for small pressure and

volume changes, we might expect that a 30% decrease in volume would result in a

30% increase in pressure.  However as we saw with the 10% change in a

preceding problem (where the pressure change differed slightly from the 10%),

and as we see here with the 30% change (where the pressure change differs more

significantly from the 30% we might have expected), when the percent change in

volume is not small, the percent change in pressure cannot be expected to be the

same.

#$&*

If the original pressure was 100 000 Pa, what are the

final pressures in these two situations?

****

The 9% decrease in pressure gives us a pressure of 91 000

Pa.

The 43% increase in pressure gives us a pressure of 143

000 Pa.

#$&*

2-bottle system

`q007.  Using a 2-bottle system, with the 'long' tube

connecting the two bottles full of water with its end under water in both tubes,

you observed that when the 'pressure tubes' and 'release tubes' are capped and

one bottle is raised the water level in that bottle decreases, while the level

in the other bottle increases.  You took data to determine the change in

volume in the lower bottle.

What were your data for this activity? 

****

Your data should have consisted of the change in water

level, the diameter or circumference of the bottle and the height to which the

first bottle was raised.

#$&*

What do you conclude was the volume change of the gas in

the bottle? 

****

Suppose the bottle has circumference 30 cm, and the water

level increased by 1.2 cm. 

The diameter of the bottle would be 30 cm / pi = 9.5 cm. 

Its radius would be half that, and its cross-sectional area about 70 cm^2.

The newly filled region of the bottle is therefore a

cylinder with cross-sectional area 70 cm^2 and altitude 1.2 cm.  Its volume

is therefore

V = A * h = 70 cm^2 * 1.2 cm = 84 cm^3.

#$&*

What is the volume change as a percent of the original

volume? 

****

Assuming the original volume of the gas in the bottle to

have been .5 liter, or 500 cm^3, the 84 cm^3 change would be

84 cm^3 / (500 cm^3) = .16 of the original volume, or 16%.

#$&*

What do you conclude is the pressure change in the bottle? 

****

The volume has decreased by 16%, to 84% of the original.

If the original pressure was P_atm, then the 16% decrease

in volume leads to a new pressure of

P1 = P_atm * (V0 / (.84 V0) ) = 1.19 P_atm.

The change in the pressure inside the bottle would

therefore be .19 P_atm = .19 * 100 000 Pa = 19 000 Pa.

#$&*

What was the height of the water column in the vertical

tube connecting the two bottles?   What is the vertical distance

between the water surfaces in the two bottles? 

****

For example, you might have raised the bottle so its water

level was 1.2 meters above that in the lower bottle, with the highest point of

the vertical tube 30 cm above that.

#$&*

What therefore must be the pressure difference between the

water surface in the lower bottle, and the water surface in the upper bottle? 

****

The pressure difference due to the 1.2 meter difference in

height is the pressure required to support a column of water 1.2 meters high. 

The pressure difference is therefore

`dP = rho g `dy = 1000 kg/m^3 * 9.8 m/s^2 * 1.2 m = 12 000

Pa, approx..

#$&*

Is this difference accounted for by the pressure change

you calculated based on the change in the water level in the lower bottle? 

****

If the lower bottle is at 19 000 Pa relative to the upper

bottle, as previously calculated, then the water level accounts for only 12 000

Pa of the difference.

It could be that we were gripping the bottle very hard. 

This could add 7000 Pa to the pressure in the upper bottle, but it's unlikely we

would grip that hard unintentionally.

Of course there's another factor.  The water that

siphoned into the lower bottle came from the upper bottle, vacating additional

space for the air in the bottle.  This doesn't help a lot, since this will

tend to lower the pressure in the upper bottle rather than raising it.

The bottom line:  Either there's 7000 Pa of

additional pressure in the upper bottle, despite the tendency for the pressure

to decrease as water is siphoned out, or uncertainties in measurement can

account for the unaccounted 7000 Pa difference.

#$&*

How would you expect the air columns in the pressure tubes

to behave in this experiment? 

****

The pressure in the lower bottle increases, which will

exert additional force on the water it the 'bottle side' of the pressure tube,

pushing the water toward the capped end of the tube.  This will decrease

the volume of the air column, which will increase the pressure and therefore

increase the force exerted on the water from this side.  The increased

pressure in the air column will thereby equalize with the pressure in the

bottle.

The pressures will also equalize in the pressure tube of

the upper bottle.  Since the volume occupied by the air in that bottle

increases as water flows from the upper bottle, this will tend to lower the

pressure in that bottle.  The air column in the pressure tube will

therefore lengthen.

#$&*

Would the air columns both get shorter, both longer, or

would one get shorter and the other longer?  Explain.

****

As indicated in preceding comments the air column in the

pressure tube of the upper bottle will get longer; the column in the lower

bottle will get shorter.

#$&*

`q008.  You did the same activity with the second

bottle 'uncapped'.  What differences did you observe and how do you

interpret them?

****

When the second bottle is uncapped its pressure doesn't

decrease when water flows from it into the lower bottle.  The pressure

difference due to the relative water levels is still the same, but since the

pressure in the upper bottle remains higher, the resulting pressure in the lower

bottle will be higher.

This will result in a lower air volume than before, in the

lower bottle.  A lower air volume means that the water level will rise

higher than before.

#$&*

Beads dropping on pan balance

`q009.  Beads of mass m_bead are dropped onto the pan

of the balance hitting the balance at velocity v and rebounding at velocity v '. 

The average time between drops is `dt.  Using symbols, reason out the

momentum change experienced by a single bead, and the average force exerted by

the dropped beads.

What is the symbolic expression for the momentum change?

****

Momentum changes from - m_bead * v to + m_bead * v ',

where we have implicitly assumed that the upward direction is positive.

The momentum change is therefore

#$&*

What is the symbolic expression for the average force?

****

#$&*

`q010.  Suppose the velocities v and v ' of the bead

in the preceding are respectively -4 m/s and 2 m/s, and that `dt is .8 second. 

Substitute these values into the symbolic expression you obtained in the

preceding.  What is your result (note that m_bead is not given, so m_bead

will remain in its symbolic form)?

****

We get

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What is the weight of a bead whose mass is m_bead?

****

The weight is

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Compare your two expressions.  How

many times the weight of a bead is the average force?

****

F_ave / wt = m_bead * 4.8 m/s^2 / (m_bead * 9.8 m/s^2) =

.49, approximately.

The average force is equal about half the weight of a

bead.

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Energy in a gas

A monatomic gas consists of atoms which behave as

particles.  Their average KE is 3/2 k T, where k = R / N_A and T is

absolute temperature. 

Since k = R / N_A this can be written as 3/2 n * N_A *

R / N_A * T, which simplifies to

3/2 n R T.

We conclude that the total kinetic energy of the

particles in n moles of a monatomic gas is 3/2 n R T.

We call this the total internal energy of the gas

and denote it by the symbol U.  We write

For a reason related to the frequent behavior of a dropped

pony bead, which usually rebounds to about 70% of its original height but which

occasionally rebounds to only a few percent of the original height before

falling back to the floor then rebounding much higher than before, a mole of

diatomic gas (air, for example, consisting predominantly of diatomic nitrogen

and diatomic oxygen and hence behaving pretty much as a diatomic gas) at a given

temperature has 5/3 as much internal energy as a mole of monatomic gas at the

same temperature.  5/3 * 3/2 n R T = 5/2 n R T, so

(we'll see the connection between this and the behavior of

the bead in our next class).

`q011.  If the temperature of n moles of monatomic

gas changes from T1 to T2, then what is the expression for the change U2 - U1 in

internal energy of the gas?

****

The internal energy goes from U1 = 3/2 n R T1 to U2 = 3/2

n R T2, a change of

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What would be the change if the gas was diatomic?

****

Internal energy at temperature T in a diatomic gas is 5/2

n R T.  Modifying the answer to the preceding we find that the change would

be

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`q012.  2 liters of confined diatomic gas in a bottle

is heated from 0 Celsius and atmospheric pressure to 50 Celsius, at constant

volume.  What is the pressure at 50 Celsius, and what is the height of a

water column that could be supported by that pressure?

****

The gas is confined so n is constant, and volume is stated

to be constant.  We conclude that P / T = n R / V remains constant, so P0 /

T0 = P1 / T1.  If P0 = P_atm, T0 = 0 Celsius = 273 Kelvin and T1 = 50

Celsius = 323 Kelvin, we have

The pressure therefore exceeds atmospheric pressure by a

factor of .18, or .18 * 100 000 Pa = 18 000 Pa.

This excess pressure will support a water column of height

rho g h, where rho g h = 18 000 Pa.  Solving for h we get

g = 18 000 Pa / (rho g)

= 18 000 Pa / (1000 kg / m^3 * 9.8 m/s^2)

= 18 000 Pa / (9800 kg / (m^2 s^2) )

= 1.83 m.

The units calculation:

Pa / (kg / (m^2 s^2) ) = (N / m^2) / (kg / m^2 s^2) = (

(kg m/s^2) / m^2 ) * (m^2 s^2 / kg) = m

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The gas is then heated to 100 Celsius, at constant

pressure.  What will be its volume at this temperature?

****

The temperature rises from 50 C to 100C, i.e., from 273 K

to 373 K. 

If n and P are constant, the V / T is constant and V0 / T0

= V1 / T1, so

The original volume was V0 = 2 liters so the new volume

will be

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The expansion of the gas displaces an amount of water

equal to the change in gas volume, to the height you found in the first part of

the question.  By how much does the PE of the system therefore increase?

****

The change in volume is .30 liters, the height at which it

flows out of the tube is 1.83 m above the level in the container.

.30 liters of water has mass .30 kg and weight .30 kg *

9.8 m/s^2 = 3 Newtons, approx..

When raised to a height of 1.83 m its PE will therefore

increase by an amount equal to the work done against gravity:

#$&*

By how much does the internal energy of the gas change as

we heat it from 0 C to 100 C?

****

The gas is diatomic so its internal energy is U = 5/2 n R

T, and the change in internal energy between absoluted temperatures T1 and T2 is

`dU = U2 - U1 = 5/2 n R ( T2 - T1), or 5/2 n R `dT.

Its volume at 273 K and atmospheric pressure was 2 liters

= .002 m^3.  Solving P V = n R T for n we obtain

The internal KE changes by

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What is the PE change of the system as a percent of the

internal energy change of the gas?

****

The PE change of the system is about 5 J, the internal

energy change is about 200 J, so the PE change is about 5/200 = .025, or 2.5%,

of the internal energy change.

#$&*

Homework:

Submit the above questions in the usual manner.

Within the next week you should master the problems in

Introductory Problem Set 5, found at

http://vhmthphy.vhcc.edu/ph2introsets/default.htm .