course phy202
Dear Mr. Smith,I fell behind for weeks when I was deployed to South Korea last month, but I am determined now to get back on track. I believe that this next test will be especially challenging, so I have decided to take you up on your offer to submit a practice test and receive feedback before I take the real thing:
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Analyze the pressure vs. volume of a bottle engine consisting of 5 liters of an ideal gas as it operates between minimum temperature 240C and max temp 300C, pumping water to half the max possible height. Sketch a pressure vs. volume graph from the original state to the max-temp 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 effiency of the process.
We are given P1=1.01E5 Pa. V1=5L. T1=513K. T2=573K. So, P2=(1.01E5*573)/513=1.13E5Pa. So, V2=(5*573)/513=5.58L. So, `dV=5.58-5=.58L. The max possible height is (1.13E5-1.01E5)/(1000*9.8)=1.22m. Half the max possible height is .5*1.22=.61m. The weight of water corresponding to `dV is .58*9.8=5.68N. The work to raise this water to half the max possible height is 5.68*.61=3.46J. n=(1.01E5*5)/(8.31*513)=118mol.
To get to half the max height you need to increase pressure without increasing volume. That requires that the system be raised to a certain temperature before it begins to expand. The amount of expansion must then be based on this temperature and the max temperature. You don't get all .58 liters of expansion because the expansion does not occur between the minimum and maximum temperatures.
Cv=5/2*8.31=20.8J/mol*K. Cp=7/2*8.31=29.1J/mol*K. Using the example in Class Notes #9, I calculate the thermal energy by averaging Cv and Cp, which gives 24.95J/mol*K. So, thermal energy required is 24.95*118*(573-513)=1.77E5J. So, efficiency is 3.46/1.77E5=1.95E-5. The graph I would sketch would resemble that shown near the top of Class Notes#10.
Cv applies to the initial pressure increase, and Cp to the expansion phase. If the temperature changes for the two phases are equal then it would be OK to average the two, but in general you want to calculate each phase separately.
If we add 13g of steam at 100 C and .9g of ice at 0 C to 190g of water at 30 C, what will be the final temp of the mixture, assuming that all thermal exchanges take place within this system? The heats of vaporization and fusion for water are about 2250J/g and 335J/g.
The energy to turn 13g of steam at 100C to water is 13*2250=29250J. The energy to turn .9g of ice to water at 0C is 335*.9=301J. The energy required to condense the steam plus the energy required to melt the ice equals the energy lost by the water, the mass of which now includes the original 190g plus the mass of condensed steam and melted ice, so 292501+301=(.19*4186*(Tf-30)+(.013*4186*(100-Tf))+(.0009*4186*(0+Tf)). So, 29551=795Tf-23860+5442-54.4Tf+0+3.77Tf. So, Tf=64.5 C.
Good.
Water is descending in a vertical pipe of diameter 8cm. At a certain level the water flows into a smaller pipe of diameter .8cm. At a certain instant the gauge pressure of the water at a point 60cm above the narrowing point is 174kPa and the water there is moving at 190cm/s. What is the gauge pressure of the water just above the narrowing point? What is the pressure change across the narrowing point?
Gauge pressure will remain the same until reaching the narrowing point. So, the gauge pressure of the water just above the narrowing point will still be 174kPa. The actual pressure is 1.74E5+1.01E5=2.75E5 Pa. Across (i.e., on the other side of) the narrowing point, the pressure is found as follows: 190=v2(.8/8)^2. So, v2=19000cm/s. 2.75E5+.5*1000*1.9^2=P2+.5*1000*190^2. So, P2=-1.77E7 Pa.
Between the given point and the point just above the narrowing point the vertical coordinate y changes while water velocity remains constant so
`dP + `d(rho g y) = 0
and you do get a pressure change.
Across the narrowing point the pressure change is as you say.
How much ice at 0 C should we add to 210g of water at 45 C and 17g of steam at 100 C to end up with only water at 0 C, assuming that all thermal exchanges take place within this system?
Energy to cool 210g of water from 45 C to 0 C= .21*4186*45=3.96E4 J. Energy to condense 17g of ice=2250*17=38250J. Energy to cool 17g of water from 100 to 0=.017*4186*100=7116 J. The sum of these is 8.50E4 J. This amount of energy must equal the energy required to melt the ice, so 8.50E4=m*335, m=254g.
good
A diatomic gas in a 1.5L container is originally at 25 C and 1atm. It is heated at constant volume until its pressure has increased by .86atm, then at constant pressure until the gas has increased its volume by .39L. 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?
We are given V1=1.5L, T1=298K, P1=1.01E5 Pa, P2=1.88E5 Pa, V2=1.89L, `dV=.39L. So, T2=(298*1.89)/1.5=375K.
This would be the temperature if the gas was first allowed to expand at constant pressure to its final volume. However if its heated first to increase its pressure.
No work is done at constant volume. The work done at constant pressure is 1.88E5*.39=1.67E6 J. n=(1.01E5*1.5)/(8.31*298)=61.2mol. Cv=5/2R=20.8J/mol*K. Cp=7/2R=29.1J/mol*K. Using the example in Class Notes #9, I calculate the thermal energy by averaging Cv and Cp, which gives 24.9J/mol*K. 24.9*61.2*(375-298)=1.17E5J. Internal energy=1.17E5-1.67E6=-1.55E6J.
You need to calculate energy for each phase and you can't use the average of Cp and Cv.
Good work overall. I believe you can correct any errors based on my notes. Feel free to submit a copy of this document, with revisions and/or questions indicated by &&&&.