#$&*
course phy 232
i didn't do number 6, the last one on the practice test. I was too tired and I don't think I knew how to go about solving it anyway. I haven't done the brief bottle experiment lab 2d yet and I think that one had something to do with that problem.
110214 Physics IIThe document
http://vhcc2.vhcc.edu/ph2spring99/classes_2003/0212.htm
consists of a practice test with notes on solutions. Test problems are in italics. You should try to work the problems without viewing. Here is a list of the problems:
Problem Number 1
Assuming that your lungs can function when under a pressure of 7.8 kPa, what is the deepest you could be under water and still breathe through a tube to the surface?
****
7.8 kPa Is 7800 N/m^2, and the change in pressure is rho*g*y, I know rho and g, so I need to know the y distance that will satisfy 7.8 kPa..so here's the setup: 7800 N/m^2 = (1000kg/m^3)(9.81 m/s^2)(X)
so X=1.26 meters
#$&*
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?
****
I know P0= 101325 Pa, I know V1=V0=7600 cm^3, and I know T0= 414 K, I also know that P1=P0 +rho*g*y, so using the ideal gas law, I can find the required temperature for a height of 99 cm.
So I know P0V0/T0=P1V1/T1, which is: T1=[(101325+(1000*9.81*.099))/101325]414
so T1 is 417.98 K, or a temperature change of 3.97 K
@& Good, if the increase in y is .099 m.
However 99 cm is 0.99 m, so the temperature change would be about 10 times what you calculated.*@
#$&*
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?
****
P0=101325 Pa, P1=188465 Pa, V1=V0=1500 cm^3, T0=298 K, using the ideal gas law, I know T1=554 K
for the part with increased volume, I know V2=1890 cm^3, using PV=nRT, for T2, I get 698 K, P2=P1=188465 Pa.
So know for the thermal energy required for both processes: for the constant volume, I get (5/2)nRT, since the gas is diatomic. The the constant pressure, I get (7/2)nRT, to account for the increasing volume. T is the change in temperature during these two processes. So for total Thermal Energy required is: [(5/2)nR(256)]+[(7/2)nR(144)], using the ideal gas law, I can solve for n, which is .061 mol. I know R is Regnault's Gas Constant of 8.31 J/(K*mol), so my new answer is: 506.9 J
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?
****
I feel strongly that I should use bernoulli's equation, since it mentions pressure, height of water, and speed of water. For this equation, lets use p=rho, and P=pressure. So I have pgy+.5pv^2+P=constant.
For the point 80cm above I get 86682.9 as the coefficient for my constant.
So for the point right above the narrowing point, I will use 0 as my height. So now I get:
.5(1000 kg/m^3)(.094m/s)^2 +P1=86682.9, so now I can solve for P1, which is 86.678 kPa
I am not sure about this one without looking it up..lets take the cross-sectional area, pi(.0006^2)m^2=
.0000011 m^2 as compared to pi(.004^2)m^2 = .00005 m^2..i know pressure is force/Area, so i'm thinking that comes in somewhere...the pressure right before it narrows is 86.678 kPa, and I know P=F/A, so I can solve for F in this case with this: 86.678 kPa=F/.00005 m^2, so the force is 4.33 N. using this force for the more narrow part, since it feels to me force would be constant throughout, so I get P2=4.33 N/ .0000011 m^2, so P2 = 3.85 MPa(that seems to large??)
@& Water is incompressible so the ratio of velocities before and after narrowing is the inverse of the ratio of cross-sectional areas.
See also the continuity equation.
Then come back to Bernoulli, which is definitely the right way to go.*@
#$&*
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 would think use the KE_trans for each different gas. So far hydrogen, it would be (3/2)nRT, which would be (3/2)(1)(8.31)(606)=7554 J..now put that into KE=.5mv^2, so I get 7554 J=.5(.002)v^2. So the velocity of 1 mole of hydrogen would be 2749 m/s
for helium, I think is monatomic, so using the same process I will get 1943 m/s
for Nitrogen, I think it is monatomic as well, so same process again. So it's speed is 735 m/s
for Oxygen, I know most of it is in the form of O2, so i'll consider it diatomic. Meaning I have KE_total=(5/2)n(8.31)(606), but only (3/2)nRT is translational, so I have energy going into rotation, which means less speed. I'm not very clear on this case, but I would think the required energy would be higher to move a diatomic molecule at the same speed as a monatomic, so a fraction below the monatomic of (3/2) makes sense...maybe the ratio of rotationalKE and translationalKE, so ((2/2)/(3/2))=2/3, so (2/3)(1)(8.31)(606) = 3357 J, so using KE=.5mv^2, I get v=458 m/s
for carbon dioxide, it's made up of 1 carbon and 2 oxygens, so wouldn't that be triatomic??? so it would have 3 degrees of translational KE as well as 3 degrees of rotational(unless the bond of all three molecules is linear, which I don't think it is)?so the energy input to output into translation ratio would be (1/2). so I get (1/2)(8.31)(606) = 2518 J
@& Good. The energy for CO2 is fairly complicated, but 3 degrees of rotational freedom is appropriate.
However C02 is not present in concentrations that significantly affect the thermal properties of air.*@
@&
Good thinking throughout. See my notes.*@