#$&*
course phy 232
I didnt do the last two because i know how to do them
Problem Number 1Assuming 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?
.5rhov+rhogH+P=.5rhov+rhogH+P
Since velocity is 0
rhogH+P=rhogH+P
Since H1= 0
101325 Kg/m/s^2=1000kg/m^3*-9.81m/s^2*H + 7800 Pa
If point A is at the surface and point B inside your lungs, the pressure at A is 1 atmosphere.
The pressure in your lungs is 1 atmosphere + rho g h + 7800 Pa.
If you set the two equal and solve I believe you get h = -.8 meters, very approximately.
H=-9.533 meters below 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?
Start with .5rhov+rhogH+P=.5rhov+rhogH+P
P= 1000kg/m^3*9.8m/s^2*.99m+101325
P= 111027 Pa
P1/T1= P2/T2
101325/141=111027/T2
T2=154.5 Celcius
The gas laws apply only to absolute temperature. Celsius is not an absolute scale. Everything else in your solution is good, so this should be easy to correct.
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?
Start off to get the change in temperature over the whole problem.
So 1/298=1.86/T2 ____ T2= 554.28
1.5/554.28=1.89/T3___T3= 698.39
1.5/554.28=1.39/T3
Find the number of moles= n=PV/RT
n= 6.057 x 10^-4
Right procedure, except that you didn't use units in your calculation and hence got the wrong number of moles. My guess is that you used volume in liters, which are different than the m^3 implicit in the units of R. You would get 6.057 * 10^-1 mole, or .06057 moles.
The internal energy of the gas changes by the equation
dQ= 5/2RndT
dQ= 5/2*8.31* 6.057 x 10^-4* 400.39
dQ= 5.038 Joules
Thermal Energy Required
Ktr= 7/2RndT
Ktr= 7/2*8.31*6.057 x 10^-4*400.39
Ktr= 7.046 Joules
Work= p(V2-V1)
W=1.86(.39)=.7254 Joules
????? I feel like I should have done something different for either the internal energy or thermal energy, are they right?????
This part is fine. Your earlier errors are easy to correct.
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?
???????There is a problem in my math somewhere do you see it??????
86600kg/m/s^2+1000kg/m^3*9.8m/s^2*.8m= P2= 94440 kg/m/s^2
V3= A2V2=A3V3= pi(.04m)^2*.94m/s^2= pi(.006m)^2*V3 V3= 41.77m/s^2 ???????? I think this is wrong but I don’t see how???????
P3=94440+.5(1000)(.94)^2=P3+.5(1000)(41.77)^2
94881.8Pa= P3 + 872366.45Pa
P3= -777484.65Pa ????? Cannot have a negative pressure?????
Water flowing fast enough through a pipe can 'pull' air in, so the negative pressure is certainly possible. I don't have time to verify all your arithmetic, but I don't see anything wrong with your thinking or your equations, and your result is very plausible. We would expect several atmospheres of negative pressure in this situation.
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 used the equation
.5mv^2=3/2kT
.5(mass in grams of element)(v^2)=3/2(1.381x10^-23)(606)
???? Is this the right k constant? I didn’t use the one in the constants page I used the one that came with the formula in the book?????
the k given there is the electrostatic constant; the value of R is given, as is Avogadro's Number, and the Boltzmann constant is k = R / N_A.
Hydrogen=1.12x10^-10
Helium=7.922x10^-11
Nitrogen=2.994x10^-11
Oxygen=2.801x10^-11
Carbon Dioxide= 2.388x10^-11
The Boltzmann constant k has units of Joules / (particle Kelvin). You need to use the mass of a particle, i.e., the mass of a single atom or molecule. That requires that you divide molar mass by Avogadro's Number.
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.
Problem Number 7
When held under tension about 170 Newtons a chain with total mass about 15 kg and length 8 meters supported a pulse which made 3 round trips in 5.4 seconds.
• How fast was the pulse moving?
• How does this compare with sqrt( tension / (mass per unit length) )?
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