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course Phy 232
6-23-11 about 7:30pmI have not finished this, but I thought I would submit what I have thus far to make sure it is right
Physics II Practice Test I
This is a practice test for Test #1 in Physics 202. Problems are in italics, hints and a couple of more detailed solutions are in regular type.
Suggested use: Work through the practice test without looking at hints/solutions, then look at hints/solutions and self-critique in the usual manner, inserting your responses into the document. Be sure to mark insertions before and after with **** so your instructor can find them quickly.
Notes for specific classes:
Physics 121 students note: Only problems generated from the Introductory Problem Sets are relevant to your test. You should be sufficiently familiar with those problems to recognize them.
Physics 201 students note: Some problem are marked for University Physics students, and you don't need to worry about them. You should of course be familiar with the Introductory Problem Sets, which comprise a significant number of problems on Test 1.
University Physics students are responsible for everything.
Constants:
k = 9*10^9 N m^2 / C^2 qE = 1.6 * 10^-19 C h = 6.63 * 10^-34 J s
energy of n=1 orbital in hydrogen atom: -13.6 eV k ' = 9 * 10^-7 T m / amp atomic mass unit: 1.66 * 10^-27 kg
electron mass: 9.11 * 10^-31 kg speed of light: 3 * 10^8 m/s Avogadro's Number: 6.023 * 10^-23 particles/mole
Gas Constant: R = 8.31 J / (mole K) proton mass: 1.6726 * 10^-27 kg neutron mass: 1.6749 * 10^-27 kg
Problem Number 1
Assuming that your lungs can function when under a pressure of 7.8 kPa, what is the deepest you could be under water and still breathe through a tube to the surface?
&&&& 100 kPa +rho*g*y_1+ ½ rho*v^2= 7.8 kPa +rho*g*y_2+ ½ rho*v^2
since v is constant v^2 factors to 1
y_1= is 0, since it is the surface of the water
solving for y_2 gives the equation . . .
y_2= (100 kPa - 7.8 kPa)/rho*g
plug in the values for rho and g and solve
y_2= 9.41m&&&&
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@& Your lungs already have 100 kPa of pressure in them, equal to the pressure of the atmosphere.
As you go deeper in the water, the pressure on your body increases. When it has increased by 7.8 kPa, your diaphragm is no longer able to exert enough force to expand your lungs, and you will no longer be able breathe in air from the atmosphere.
So y_2 would be just -7.8 kPa / rho g, and the depth would be about .8 meters.*@
Bernoulli's Eqn applies, with v presumed constant.
Problem Number 2
There is a small amount of water at the bottom of a sealed container of volume 7.6 liters which is otherwise full of an ideal gas. A thin tube open to the atmosphere extends down into the water, and up to a height of 143 cm. The system is initially at atmospheric pressure and temperature 141 Celsius.
If we increase the temperature of the gas until water rises in the tube to a height of 99 cm, then what is the temperature at that instant?
&&&& solve for p_2 in Bernoullis equation
p_2= p_1- ½ rho - rho*g(y_2 - y_1)
@& p_2= p_1- ½ rho (v_2^2 - v_1^2) - rho*g(y_2 - y_1)
v_1 = v_2 = 0 in this case so it won't make a difference, but in general you don't want to leave the v out of the 1/2 rho v^2 terms.
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y_1=0 since no water is in the tube initially
p_1= 100 kPa
plugging in known values and solving gives p_2= 89.798 kPa
use PV=nRT, get all constants on one side
p_1/T_1= p_2/T_2
100 kPa/414 K= 89.798 kPa/T_2
241.5 kPa/ K= 89.798 kPa/T_2
T_2= 371.8 K
I know this is wrong since the temp I ended up with is lower than the initial temp of 414 K. Im not sure what Ive done wrong.&&&&
@& Your instincts are correct. The pressure in the bottle will of course be higher than atmospheric pressure, not lower.
You need to carefully consider which point is #1 and which point is #2:
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@& p_2= p_1- ½ rho (v_2^2 - v_1^2) - rho*g(y_2 - y_1)
Atmospheric pressure would occur at the top of the tube, so if P_2 is the pressure in the gas, y_2 would be lower than y_1. This would make y_2 - y_1 negative, and would giver you P_2 = 110 kPa, approx.
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Bernoulli's Equation will give us the new pressure.
If the tube is thin then volume change is negligible so PV = nRT tells us that P / T is constant.
From init pressure, final pressure, init temp we easily find final temp.
Problem Number 3
A diatomic gas in a 1.5-liter container is originally at 25 Celsius and atmospheric pressure. It is heated at constant volume until its pressure has increased by .86 atm, then at constant pressure until the gas has increased its volume by .39 liters. How much thermal energy is required? By how much does the internal energy of the gas change? How much work is done in the process?
&&&&for state 1: V=1.5liters, T=298 K, P= 100 kPa
State 2: V=1.5 liters, T= (P_2/P_1)T_1= 554.3 K, P= 186 kPa
State 3: V= 1.89 liters, T= (V_1/T_1)(1/V_2)= 700 K, P= 186 kPa
State 1 height= P/(rho*g)=10.2cm
@& `dP + rho g `dh = 0 in this case, since velocity is low and 1/2 rho v^2 can therefore be ignored.
However P is not equal to rho g h. The equation is written in terms of changes in P, rho g h and 1/2 rho v^2.
So you wouldn'g divide P by rho g to get h.*@
State 2 h= 18.98cm
State 3 h= Im not sure how to find this
State 1 KE= 14.7N*.102m= 1.5J, F= 1.5kg*9.8m/s^2= 14.7N
State 2 KE= 18.5N*.1898m= 3.5J, F= 1.89kg*9.8m/s^2= 18.5N
State 3 KE= ?
This is what Ive got so far&&&&
@& There is in any case no assumption that anything is changing altitude. In some systems the increase in pressure is associated with water rising in a thin tube, in which case the water would rise to a height of about 8.4 meters, and while that isn't relevant to this situaiton it's good to know. Be sure you know how to get this result in case you need it.
The questions you need to answer are:
What is the specific heat of the gas?
How many moles of gas are present?
How much energy is therefore required to heat the gas?
The specific heat of a gas at constant volume is 3/2 R (ideal monatormic gas) or 5/2 R (ideal diatomic gas), where R is the gas constant from P V = n R T. You should know, based on kinetic theory, why this is so. This theory, as presented in the introductory problem sets, starts with the model of a single mass bouncing back and forth between the walls of a container and ends with P V = n R T.
From the given information on pressure and volume you can find n.
At constant pressure the specific heat is R greater greater than at constant volume, because of the work dong in expanding the gas against the pressure.
Actually as I look lower in this document I see addiitional information has been included. Can you give me your thinking about the results indicated by the question marks?
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The rough sketch below is helpful in assembling the information given in this problem.
First state:
V = 1.5 L
T = 298 K
P = Patm
Second state:
V = 1.5 L
T = T1 = ?
P = Patm + .89 atm
V const so P/T is const.
We easily find T1.
Third state:
V = 1.89 L
T = T2 = ?
P = Patm + .89 atm
P const so V / T const.
We easily find T2.
Thermal energy changes:
3/2 k T per particle for translational KE, or 3/2 n R T for entire system, implies 1/2 k T and 1/2 n R T, respectively, for each degree of freedom.
Diatomic molecule has two degrees of rotational freedom, so total for three degrees of translational and two degrees of rotational freedom is 5/2 n R T.
If we know n and `dT we easily find the thermal energy changes between the two states.
n = P V / ( R T) for any of the states of the system.
For expansion we again have to increase internal KE, which is still 5/2 n R T.
We also have to do the work of the expansion, which at constant pressure is 2/2 n R T.
So thermal energy required to expand at const P is 7/2 n R T.
More about energy changes in general
`dQ = 5/2 n R `dT is thermal energy required to raise temp of diatomic gas by temp change `dT.
This is of the form `dQ = k m `dT for specific heats, except instead of m we have n and instead of spec heat k we have 5/2 R.
5/2 R = 20 J / (mol K), approx., is the molar specific heat. Just as we multiply number of J / (kg C) by # of kg and change in Celsius temp to get thermal energy change, here we multiply J/(mol K) by # moles and change in K to get thermal energy change.
This is the thermal energy required to raise temp of a gas at constant volume.
We say Cv = 5/2 R, and `dQ = Cv * n * `dT, where Cv is molar specific heat at constant volume.
At constant pressure we see from the above that Cp, the molar specific heat at constant pressure, is 7/2 R.
The ratio Cp / Cv, which for a diatomic gas is
`gamma = 7/2 R / (5/2 R) = 1.4.
For monatomic gases (only 3 degrees of freedom) we have Cv = 3/2 n R T and Cp = 5/2 n R T, so for a monatomic gas
gamma = 5/2 N / (3/2 R) = 5/3 = 1.67 approx.
Note that all this is for ideal gases.
Adiabatic expansion or compression of a gas:
P V^`gamma = constant
Problem Number 4
Water is descending in a vertical pipe of diameter 8 cm. At a certain level the water flows into a smaller pipe of diameter 1.2 cm. At a certain instant the gauge pressure of the water at a point 80 cm above the narrowing point is 86.6 kPa and the water there is moving at 94 cm/s. What is the gauge pressure of the water just above the narrowing point? What is the pressure change across the narrowing point?
Three points are involved here, the 'known point' 80 cm above, the point just above, and the point just below the narrowing. Number these points 0, 1 and 2. We will apply Bernoulli's Equation to these points, two points at a time..
We're given P0, y0 and v0. So we know the required variables for State 0. We will compare each of the other points to point 0.
At point 1 we know y1 and v1 (v1 = v0). So we easily find `dP and therefore P1.
At point 2 we know y2, from dimensions we can find v2 so again we easily find `dP and therefore P2.
Problem Number 5
The masses of 1 mole of various gases are as follows: hydrogen about 2 grams, helium about 4 grams, nitrogen about 28 grams, oxygen about 32 grams and carbon dioxide about 44 grams. On the average how fast does a molecule of each gas move at 333 Celsius?
Ave KE per particle is KEave = 3/2 k T.
It follows that .5 m v^2 = 3/2 k T. We easily find v if we just know m.
We know how many grams per mole so we easily find the number of grams per particle.
Problem Number 6
Explain how to use energy considerations to determine the velocity with which water will flow from a hole in a large container if the pressure difference between the inside and outside of the container is 3100 N/m^2, and if the water inside the container is effectively stationary. You may do this symbolically or you may consider the energy changes as a 1-gram mass of water exits the cylinder.
Consider a 'plug' of water with cs area A and length L.
One end of the 'plug' is in the container and experiences the pressure in the container. The other end is exposed to atmospheric pressure.
We can obtain expressions for the net force on the plug, then for the work done in pushing the plug out, so we know the KE of the plug as it exits.
The expression for the volume and therefore the expression for the mass of the plug are easily found.
Setting .5 m v^2 = KE we easily find the expression for v.
University Physics Problem: For a gas, density is a function of pressure and temperature. Assuming that temperatures do not vary significantly, express Bernoullli's Equation in differential form, with density, pressure, velocity and altitude all variable. (Hint: Start with `d(`rho g h) + `d(.5 `rho v^2) + `dP = 0).
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@& I've inserted a number of notes. I'll look forward to responses, questions, etc., and solutions to the remaining problems.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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