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course PHY 202
2/17/15 at 11:15PM
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?
****P2-P1 = 7800 Pa
P1 + pgh1 + 1/2pv1^2 = P2 + pgh2 + 1/2pv2^2
P1 + (1000kg/m^3)(9.8m/s/s)(0m) + 1/2(1000kg/m^3)(0m/s)^2 = P2 + (1000kg/m^3)(9.8m/s/s)h2 + 1/2(1000kg/m^3)(0m/s)^2
-(1000)(9.8)h2 = P2-P1
-9800h2 = 7800 Pa
h2 = -0.8m
Thus, you can be 0.8m underwater and still breathe through a tube to the surface.****
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?
**** P1 + pgh1 + 1/2pv1^2 = P2 + pgh2 + 1/2pv2^2
(101300Pa) + (1000kg/m^3)(9.8m/s/s)(0m) + 1/2(1000kg/m^3)(0m/s)^2 = P2 + (1000kg/m^3)(9.8m/s/s)(0.99m) + 1/2(1000kg/m^3)(0m/s)^2
101300 Pa = P2 + 9702
P2 = 91598 Pa
P1/T1 = P2/T2
101300/(141+273K) = 91598/T2
T2 = 374 K = 101 C ****
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It's the pressure at the top of the open tube that is equal to atmospheric pressure.
This leads you to a pressure of about 110 kPa, and a final temperature higher than the initial temperature .
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The pressure in the bottle is higher than that of the atmosphere. This is intuitively obvious.
It also follows from careful use of Bernoulli's equation. As you descend from the top of the water column to the water level in the bottle, altitude y decreases and, consistent with Bernoulli's equation, pressure therefore increases.
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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?
**** I initially sketched the heat cycle out on a graph and converted atm to Pa and L to m^3. I knew that the work done in the cycle was equivalent to the area enclosed by it.
1 atm = 101300 Pa
1.86 atm = 188418 Pa
1.5 L = 1.5*10^-6 m^3
1.89 L = 1.89*10^-6 m^3
W = (188418 - 101300 Pa)*(1.89*10^-6 - 1.5*10^-6 m^3)
W = 0.03398 J
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Work at constant pressure it equal to the product of the pressure and the change in volume.
`dW = P `dV
Your solution assumes that `dW = `dP * `dV.
The volume changes at constant pressure 188 kPa.
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Also, be careful with your conversions. A m^3 corresponds to 1000 liters, not 1 000 000 liters.
Your final result should be around 7 Joules.
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(This seemed small to me, and I wasn’t sure how to continue with regards to thermal energy, so I looked at the hints at this point.)
I hadn’t been thinking about the three states as systematically as I should have been, so I outlined them as you did with the information given in the problem. From there it immediately made sense how to find the missing link, T2.
State 1 has V1 = 1.5*10^-6 m^3, T1 = 298 K, P1 = 101300 Pa
State 2 has V2 = 1.5*10^-6 m^3, T2 = ?, P2 = 188418 Pa
State 3 has V3 = 1.89*10^-6 m^3, T3 = T2, P3 = 188418 Pa
P1/T1 = P2/T2
101300/298 = 188418/T2
T2 = 554.3 K
dQ for transition from 1 to 2 = 5/2nRT
where n = PV/RT = [(101300)(1.5*10^-6)] / [(8.31)(298)]
n = 6.14*10^-5 mol
(This value seems a little low to me, but I could be wrong).
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Again it's the m^3 to liters conversion that gives you the low result.
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Regardless, n should stay the same throughout because there is no mention of adding or removing any gas from the system. I would have thought that 1.5L / 22.4 L at STP would have also given n, since 1 mole at STP is 22.4 L, but this value does not equal the above value, so I’ll stick with the above. Can you explain why 1.5/22.4 is incorrect????
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The initial state was close to, but not quite the same as, STP.
If you plug 273 K, 101.3 kPa and 1 mole into PV = nRT and solve for V, you will get .0224 m^3 = 22.4 liters.
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In this case, the result you get should therefore be close to, but not quite the same as 1.5 L / 22.5 (L/mole) = .07 mole, approx.
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dQ = 5/2 (6.14*10-5)(8.31)(554.3-298)
dQ = 0.327 J for State 1 to State 2
Since there is a dT = 0 for State 2 to State 3 transition, there is no thermal energy exchange.
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You can't expand a fixed number of moles at constant pressure without heating the gas. `dT cannot be zero for this phase.
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Note that no thermal energy exchange does not mean that `dU = 0. A gas can do work at the expense of internal energy, and if work is done on the gas without the exchange of heat, the internal energy of the gas will increase.
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dU = Q - W
= 0.327 J - 0.03398 J
dU = 0.29302 J
(All of these values seem awfully low to me.)****
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It's the m^3 vs. liters thing. Your values are indeed all low, by a factor of 1000.
It's very good that you are comparing your answers to common sense.
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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?
****My approach was slightly different than yours, and when I was unsure about the validity of my values, I checked your hints, did the problem the way you outlined it and still received the same result (so I hope that means that my answer is correct). Then I realized that I had missed the first question about the gauge pressure above the narrowing point.
A1v1 = A2v2
Pi*(0.04m)^2(0.94m/s) = pi*(0.006m)^2(v2)
v2 = 41.8 m/s
P1 + pgh1 + 1/2pv1^2 = P2 + pgh2 + 1/2pv2^2
86600Pa + (1000kg/m^3)(9.8m/s/s)(0.8m) + 1/2(1000kg/m^3)(0.94m/s)^2 = P2 + (1000kg/m^3)(9.8m/s/s)(0m) + 1/2(1000kg/m^3)(41.8m/s)^2
P2 = -778738.2 Pa
The pressure after the narrowing point is -778738.2 Pa, which seems very low, but it should be lower than the gauge pressure higher up the tube due to the narrower diameter.
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Correct.
The pressure is in fact so low that air will be forced into the pipe by the higher pressure outside.
When this happens, things get complicated because the volume of the stuff below the opening will exceed that above, and the air will be compressed or decompressed by the pressure changes. That would take us beyond the scope of this course.
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I wasn’t sure what height to use for the point above the narrowing, since the point below is 0m, so I used 0.01m to be just above that point.
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In this case, as in most, the difference in altitude across the narrowing has very little effect and, after checking to see that it is so, can be considered negligible.
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P0 + pgh0 + 1/2pv0^2 = P1 + pgh1 + 1/2pv1^2
86600Pa + (1000kg/m^3)(9.8m/s/s)(0.8m) + 1/2(1000kg/m^3)(0.94m/s)^2 = P2 + (1000kg/m^3)(9.8m/s/s)(0.01m) + 1/2(1000kg/m^3)(0.94m/s)^2
P1 = 94342 Pa
Thus the pressure difference across the narrowing point is -778738.2Pa - 94342Pa = -873080.2Pa.****
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?
****I admit that I did have to look at the hints for this one because I could not recall the equation.
H2: 3/2(9*10^9)(333+273) = 1/2(2g/6.022*10^23)v^2
H2: v = 2.22*10^18 m/s
He: 3/2(9*10^9)(333+273) = 1/2(4g/6.022*10^23)v^2
He: v = 1.57*10^18 m/s
N2: 3/2(9*10^9)(333+273) = 1/2(28g/6.022*10^23)v^2
N2: v = 5.93*10^17 m/s
O2: 3/2(9*10^9)(333+273) = 1/2(32g/6.022*10^23)v^2
O2: v = 5.55*10^17 m/s
CO2: 3/2(9*10^9)(333+273) = 1/2(44g/6.022*10^23)v^2
CO2: v = 4.73*10^17 m/s****
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.
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k in this case is the gas constant, not the electrostatic constant.
The units of the latter would not work here.
Based on your understanding of kinetic theory you would be expected to know how k and R are related, with k = R / N_A (k is about 1.38 * 10^-23 J / (particle Kelvin).
If you use the correct value of k, you will get correct results for the particle velocities.
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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.
****m = 1g = Vp
V = m/p = 0.001kg/1000kg/m^3
V = 1*10^-6 m^3
Net force on plug = F_inside - F_atm
Where F = P*A
Net force = P_in*A - P_atm*A
= A(P_in - P_atm)
= A(3100N/m^2)
W = Fd = 3100A*d
d is length of plug, so A*d = V = 1*10^-6 m^3
so W = (3100N/m^2)(1*10^-6m^3)
KE = 1/2mv^2
(3100)(1*10^-6) = 1/2(0.001kg)v^2
v = 2.5m/s
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Good.
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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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You appear to be in pretty good shape.
Check my notes and let me know if you have additional questions.
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Good luck on the test.
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