course Phy 232
I'm pretty sure I did the majority of these problems wrong. I left my book and my notebook at home, so I couldn't really look for a way to do some of the problems.
Problem Number 1 Assuming that your lungs can function when under a pressure of 8.3 kPa, what is the deepest you could be under water and still breathe through a tube to the surface?
Solution: To solve this problem I first converted the 8.3 kPa to mm Hg which would be 62.255 mm Hg. I then found p in the equation d=p/w*g by taking 62.255/760 = 8191 N/m^2. I then could found the depth by taking 8191/(1000*9.8) = 0.836 meters.
Ok, but no need to convert to mm of hg. Just set rho g h equal to 8300 N/m^2, which is 8.3 kPa, and solve for h.
Problem Number 2
Find the average pressure exerted by 870 particles each of mass .011 kg traveling always at a speed of 150 m/s on one end wall of a cylindrical container of length 1.5 meters and cross-sectional area .47 m^2, provided the particle travels always along the axis of the cylinder and collides elastically with the ends of the container.
Solution: I solved this problem by first taking 870 *0.011= 9.57 which is the mass. I then took this times the velocity which would be 9.57 *150 = 1435.5. I then think this would be the force, but I am a little confused so I just took 1435.5/0.47 = 3054 for the pressure. This seems like too much, but this is what I got when I solved it I am also not sure of the units.
The force on a wall is the result of the momentum change, per unit of time, at that wall. This is according to the impulse-momentum theorem.
The total momentum of the particles corresponds to your 9.57. However you need units. The total momentum is 9.57 kg m/s.
The change in momentum at the end of the cylinder is double the momentum, since the collision is elastic. If you divide the change in momentum by the time required for a round trip, you get the result.
This problem is covered in the assigned Introductory Problem Sets, which you should understand thoroughly.
Problem Number 3
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. At what temperature will each type of gas have an average (rms) molecular speed of 313 meters / sec?
Solution: To first solve this problem I used the equation Yrms=sqrt(3*R*T/M) where M = molar mass. I solved for T in this equation which gave me T=Y^2*M/3*R. For each problem M would be different, but everything else would stay the same even R because it is a constant= 8.31J/mole K. So for hydrogen we would convert it to 1.94 moles H which is M. We would then plug it in and get approx. 7623 which is absolutely ridiculous so I must have messed up on some step. I think it is somewhere with the units. Anyways for the other elements I got 3925 for He, 7820 for Nitrogen, 7859 for Oxygen, and finally 3925 for CO^2. The units would be K. I think I may have missed a decimal place or something.
The molar mass is 1.4 grams.
If you use units properly you will get a temperature of about 8 degrees Kelving.
Problem Number 4
Find an expression for the average pressure exerted by N identical particles each of mass m traveling always at speed v on one end wall of a cylindrical container of length L meters and cross-sectional area A m^2, provided the particle travels always along the axis of the cylinder and collides elastically with the ends of the container.
• Use your expression to show whether the pressure is proportional to the velocity v of the particle, inversely proportional to v, proportional to v^2 or inversely proportional to v^2.
• Show whether the pressure is proportional to the KE of the particle, inversely proportional to this KE, proportional to KE^2 or inversely proportional to KE^2.
Solution: I just did a problem like this, but I am really unsure how to set up the formula. I know from previous experience that pressure and velocity are inversely related P=1/V. I’m not to sure about the kinetic energy, but my guess would also be inversely proportional.
Check out the Introductory Problem Sets.
Problem Number 5
Analyze the pressure vs. volume of a 'bottle engine' consisting of 7 liters of an ideal gas as it operates between minimum temperature 220 Celsius and maximum temperature 410 Celsius, pumping water to half the maximum possible height. Sketch a pressure vs. volume graph from the original state to the maximum-temperature state and use the graph to determine the useful work done by the expansion. Then, assuming a diatomic gas, determine the thermal energy required to perform the work and the resulting practical efficiency of the process.
• University Physics Students: Analyze the entire process, assuming that after the maximum temperature is achieved pressure is suddenly released, resulting in an adiabatic expansion.
• Repeat for minimum and maximum temperatures 220 Celsius and 600 Celsius. Compare the graphs you obtain and explain why the work done does not increase in proportion to the increased temperature difference.
This problem assumes you have done the labs with the bottle and tubes, and that you have viewed and taken notes on the experiments that were to be viewed.
The Class NOtes also address the situation.
Review all those sources then get back to me on this one and I'll be glad to help you clear it up.
Solution: I think the graph would look something like as the temperature increased the volume would go down and the pressure would rise, but I’m not positive on that. I believe in a adiabatic expansion everything would remain constant because this means that there has been no loss or gain of heat. I’m not exactly sure on the second question on the work done. This may be because I don’t believe I am really seeing what the graph should look like.
Part of the graph would behave in the way you describe.
For an adiabatic expansion P V^gamma is constant (meaning it stays equal to whatever it was at the start of the expansion).
For a constant-volume expansion the graph is vertical.
For a constant-pressure expansion the graph is horizontal.
The maximum possible temperature ratio is T_max / T_min. When multiplied by the original pressure, you get the maximum possible pressure. From the difference of these pressures you get the maximum possible height of the water column.
The engine works between zero height and half the max height of the water column, as stated. You will see a constant-volume rise in the graph, followed by a constant-pressure run, which ends when T_max is reached.
Problem Number 6
A tube 3.8 mm in diameter is run through the stopper of a sealed 1.5-liter container. The tube outside the container forms a U, then runs in a straight line with slope .021 with respect to horizontal. Alcohol is introduced into the tube, and fills the U, extending into the linear section of the tube. Both ends of the tube are open. The container is slightly heated, and the alcohol column is observed to move along the linear section of the tube. The material of which the container is constructed has coefficient of linear expansion `alpha = 21 * 10^6 / C. If the temperature of the air in the container was originally 29 Celsius, then if the temperature increases by .28 Celsius how far will the alcohol column move
The tube is open, so the gas in the system expands at constant pressure. If the tube is horizontal, the change in volume will be equal to the 'new' volume occupied by fluid in the tube, as the water column moves from its original to its final position.
If the tube is vertical, then any increase in the fluid level in the tube is associated with a pressure increase in the system. Assuming a thin tube, there will be negligible expansion.
If the tube is close to horizontal, as in this case, it will take a significant amount of additional fluid in the tube to increase the vertical level of the fluid, and there will be significant expansion as well as an increase in pressure.
Class Notes include a situation of this nature.
Problem Number 7
A certain material has density 6.6 kg / liter. If .26 kg of the material are suspended from a string and immersed in a liquid whose density is .85 kg / m^3, what will be the tension in the string?
Solution: I couldn’t find a formula for tension. I must be overlooking it somewhere I thought it was in my notes. So I basically have to guess at this problem. I converted the 6.6 kg/liter to kg/m^3 which would be 660 kg/m^3. I then multiplied .26 and then took that answer and divided by .85. My answer for this problem was 201, but I am pretty sure this is incorrect.
A cubic meter is 1000 liters, so the density would be 6600 kg / m^3.
This allows you to figure out the mass of the object.
The buoyant force of the liquid is equal to the liquid mass displaced by the object.
The object is in equilibrium, so the sum of its weight, the buoyant force and the tension of the string must be zero.
Problem Number 8
University Physics Problem: Use Bernoulli's equation to determine the velocity with which water will exit from a hole in a uniform cylinder when the cylinder is filled to a point 4.4 meters above the hole, assuming that the water in the cylinder moves with negligible velocity. If the cylinder has a circular cross-section with radius 5 meters while the hole has radius .051 meters, then what is the total kinetic energy of the water in the cylinder above the hole? Describe what would happen if the hole was suddenly plugged. Explain why, if the hole is instantly unplugged, the exiting water requires a short time interval to reach its maximum velocity.
Solution: Another problem where I was unable to locate the formula for this especially a Bernoulli equation that involves KE. I’m not real sure what to do on this problem. If the hole was suddenly closed then the water flow would seize. The exiting water needs a short interval to reach maximum velocity because it needs the water flow to be backing it which increases the velocity.
Bernoulli's equation will give you the exit velocity of the water. Combined with the cross-sectional area of the exit hole (which you can easily find from its given radius) you can figure out the rate at which volume flows out of the cylinder. From this rate, you can figure out how fast the water in the cylinder is descending. This velocity of descent, along with the mass of the water in the cylinder, gives you its kinetic energy.
This same volume is
Problem Number 9
Use Bernoulli's equation to determine the pressure change as water flows through a full pipe from a point where the pipe diameter is .67 meters and velocity 1.3 m/s to a point 4.1 meters lower where the pipe diameter is .4288 meters.
Solution: I must have wrote the wrong formula for Bernoulli’s Equation. I think to find the velocity for the other side you would take .67 *1.3 = 0.871 then divide 0.4288 = 2.03 m/s then to find the pressure I guess you could put 1/ 2.03 = 0.493 would be the pressure difference.
There will be a diffrence in velocities between the two points, as well as a difference in altitude. You can assume that both points are open to the atmosphere so there will be no pressure difference.
Bernoulli's equation therefore holds.
The ratio of the velocities is inversely proportional to the ratio of thde cross-sectional areas.
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