#$&*
course PHY 202
3/29 about 5pm
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?
****
‘rho*g*h1 + P1 = ‘rho*g*h2 + P2, h1 = 0, P1 = P_atm, P2 = P_atm + 7800 Pa
0 + 101325 Pa = 1000 kg/m^3 * 9.8 m/s^2 * h2 + (101325 Pa +7800 Pa)
H2 = -0.796 m
****
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?
****
‘rho*g*h1 + P1 = ‘rho*g*h2 + P2, h1 = 0.99 m, P1 = P_atm, h2 = 0 m
‘rho*g*0.99 m + 101300 Pa = P2 = 111,002 Pa
If volume is assumed constant, P1/T1 = P2/T2, T2 = 426 C
****
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?
The rough sketch below is helpful in assembling the information given in this problem.
****
V1 =1.5 L * 1000 ml/L * (1 m)^3/(100 cm)^3 = 0.0015 m^3
T1 = 298 K
P1 = 101325 Pa
V2 = 1.5 L * 1000 ml/L * (1 m)^3/(100 cm)^3 = 0.0015 m^3
T2 = (188465 Pa * 298 K) / 101325 Pa = 554 K
P2 = 101325 Pa * 1.86 = 188465 Pa
V3 = 1.89 L * 1000 ml/L * (1 m)^3/(100 cm)^3 = 0.00189 m^3
T3 = (554 K * 0.00189 m^3) / 0.0015 m^3 = 698 K
P3 = 188465 Pa
‘dQ = (5/2)*n*R*’dT + (7/2)*n*R*’dT, The first portion is the first phase with change in pressure and the second portion is the second phase with constant volume. I need to find n first.
n = PV/RT = (101325 N/m^2 * 0.0015 m^3)/(8.31 J/mol*K * 298 K) = 0.061 moles
‘dQ = (5/2)*0.061 moles * 8.31 J/K*mol * (554 K - 298 K) + (7/2)*0.061 moles * 8.31 J/K*mol * (698 K - 554 K) = 579.905 J or approx. 580 J
The work done by the system is ‘dW = P*’dV only for the second phase from stage 2 to stage 3. ‘dW = 188465 Pa * 0.00039 m^3) = 73.5 J
‘dU = ‘dQ - ‘dW = 580 J - 73.5 J = 506.5 J
****
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?
H1 = .8 m
V1 = 0.94 m/s
D1 = 0.08 m
P1 = 86,600 Pa
H2 = 0 m
V2 = 0.94 m/s
D2 = 0.08 m
P2 = ?
H3 = 0 m
V3 = ?
D3 = 0.012 m
P3 = ?
V3 = (d1/d2)^2*v2 = (0.8m)^2 / (0.012 m)^2 * 0.94 m/s = 41.78 m/s
For the pressure just above the narrowing point, point 2, velocity is constant and h2 = 0
‘rho*g*h1 + P1 = P2 = 1000 kg/m^3 * 9.8 m/s^2 * 0.8 m + 86,600 Pa = 94,400 Pa = P2
For the point just below the narrowing point, point 3, h is essentially constant.
.5*’rho*v1^2 + P2 = .5*’rho*v2^2 + P3
.5*1000 kg/m^3*(0.94 m/s)^2 + 94,440 Pa = .5*1000 kg/m^3 * (41.78 m/s)^2 + P3
P3 = -777,902 Pa
P3-P2 = -777,902 Pa - 86,600 Pa = -864,502 Pa
This doesn’t seem right but I double checked it so maybe it is since the velocity increases so much.
@&
That negative pressure will tend to draw air in from outside.
*@
****
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?
****
To find the mass, m = mass of one mole / N_A
Hydrogen - 0.002 kg/mole / 6.023*10^23 mol^-1 = 3.3*10^27 kg
.5*m*v^2 = (3/2)*k*T solved for v, v = ‘sqrt((3*k*T)/m)
V_hydrogen = ‘sqrt((3*(8.31 J/mole*k / 6.023*10^23 mol^-1)*606 K)/(3.3*10^27 kg) = 2743.59 m/s
M_helium = 6.6*10^-27 kg, v_helium = 1940.01 m/s
M_Nitrogen = 4.7*10^-26 kg, v_nitrogen = 726.987 m/s
M_oxygen = 5.3*10^-26 kg, v_oxygen = 684.602 m/s
M_CO2 = 7.3*10^-26, v_CO2 = 583.33 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.
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.
****
My first thought was to set KE = -PE, so .5*m*v^2 = -(-mgh), solve for v and simplify to get v=’sqrt(2*g*h). ????I understand the below solution, but would this not give the same answer?????
For below, you would find the force from the absolute pressure and divide by the cross sectional area. Multiplying by L would give you W. set this equal to .5*m*v^2 and solve for v.
I think I answered my own question above since we are not given h.
@&
Both solutions do give you the same result.
*@
****
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.
@&
You appear to be in good shape.
I'll be glad to answer additional questions if you have them.
*@