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course PHY 232
3:00 pm 2/22
q001. If the air in a bottle builds pressure at constant volume from temperature 300 K to temperature 450 K, then expands at this constant pressure from 450 K to 900 K:What is the ratio of highest pressure to the lowest?
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Pressure only increases in the first phase so the `d T_01 from then is the only `d T that affects P.
P_H=P_L*450/300 = 3/2P_L
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What is the ratio of the highest volume to the lowest?
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Here only `d T_12 affects V so
V_H= 900/450*V_L = 2*V_L
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In terms of the initial gas volume and pressure V_0 and P_0:
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The above need only be modified by subbing in V_L = V_0 and likewise P_L=P_0.
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What are the volume and pressure when the system reaches 450 K?
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V_450= V_0
P_450= 3/2 P_0
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What are the volume and pressure when the system reaches 900 K?
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P_900= P_450= 3/2 P_0
V_900 =2*V_450=2*V_0
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If the added pressure is used to raise water, how high will it be raised?
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Bernoulli:
P_450 = P_0+ rho*g*y
P_450= 3/2 P_0
Thus
y= 0.5P_0/(rho*g)
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If the added gas volume displaces water to the preceding height, what volume of water will be raised?
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At first I wrote:
V= A_cs*y= A_cs*0.5P_0/(rho*g)
A_cs= cross-sectional area of tube
but it occurs to me after reading the next question that what was intended was to say that the volume of water displaced will be equal to the expansion of the gas which was a net change of 2*V_0 but the change will be `d V = 2V_0-V_0 = V_0where V_0 is init volume of gas and not the water.
@& To the extent that the tube is thin, it doesn't contain a significant mass of water, and the water in it has no significant PE relative to its original position.*@
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What do we get if we multiply the volume of water raised by the height to which it is raised?
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`d V*y= V_0(0.5P_0/(rho*g)) = 1/2P_0V_0(1/rho*g)
Now this is similar to the `d PE expression we had sometime ago that was:
`d PE= P_0V_0(Ti/Tc-1)(Th/Ti-1)
Here Ti/Tc= 450/300= 1/5 and Th/Ti= 900//450= 2 which makes the eqn:
`d PE = P_0V_0(Ti/Tc-1)(Th/Ti-1) = P_0V_0(3/2-1)(2-1) = 1/2P_0V_0
So all we have to do is multiply our original eqn by rho*g and that gives us the `d PE of the system. Maybe I missed some part of the process that should have already taken care of that rho*g but it's easy enough to get rid.
In symbols:
`d PE/(rho*g) = 1/2P_0V_0(1/rho*g)
`d PE = 1/2P_0V_0
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Answer the same sequence of questions if the pressure builds at constant volume from 300 K to 600 K, then the gas is allowed to expand at this constant pressure until the temperature reaches 900 K.
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I think I just did this since I used the values of T that were given to begin with.
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Now we replace the specific temperatures by symbols.
If the temperature starts at T_c and is raised to T_i while the volume remains constant, what will be the new pressure?
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Here I'll use the phase notation of 01 and 12.
P_1= P_0*T_i/T_c
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If the temperature is then raised from T_i to T_h while the pressure remains constant, what will be the new volume?
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V_1=V_0
V_2= V_0*T_h/T_i
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What is the expression for the change in the pressure?
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`d P= P_1-P_0= P_0(T_i/T_c) - P_0
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What is the expression for the change in the volume?
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`d V = V_2 - V_0 = V_0(T_h/T_i) - V_0
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What do we get when we multiply the change in pressure by the change in volume?
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`d P* `d V= (P_0(T_i/T_c) - P_0)*(V_0(T_h/T_i) - V_0) = P_0V_0((T_i/T_c)-1)((T_h/T_i-1)
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What does our product have to do with the PE gain of the system?
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It looks like it is `d PE. I used the expression we derived some time ago above and it is this exact expression just above. Now when I did this with the values of T we had a rho*g to do eliminate but that was easy.
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What would we have to do with our expression to get the actual PE gain?
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This question makes me question my expression I obtained above since I arrived at the `d PE gain right away. I must mentioned that with the actual values of T I had a rho*g to take care of but in that process I used y in terms of P_0 whereas here I skipped that step and I think that's what made the difference between what I had earlier and what I arrived at here.
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@& Pressure changes from P_0 to (T_i / T_c) * P_0, a change of
`dPE = T_i / T_c P_0 - P_0 = (T_i / T_c - 1) * P_0.
Volume changes from V_0 to (T_h / T_i) * V_0, a change of
`dV = (T_h / T_i) * V_0 - V_0 = (T_h - T_i - 1) V_0.
Water would be raise to height y such that
rho g y = `dP = P_0
so that
y = (T_i - T_c - 1) P_0 / (rho g).
The mass of water raised would be
m = `dV * rho
its weight would be
m g = `dV * rho g = (T_h / T_i - 1) V_0 * rho g
and its PE increase would be
`dPE = weight * y = (T_h / T_i - 1) V_0 * rho g * (T_h - T_i - 1) P_0 / (rho g) = P_0 V_0 (T_h / T_i - 1) (T_h - T_i - 1)
This is the same as the product of `dV and `dP:
`dV * `dP = (T_i - T_c - 1) P_0 * (T_h - T_i - 1) * V_0 = P_0 V_0 (T_h / T_i - 1) (T_h - T_i - 1)
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@& The two expressions are the same, as explained here.
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Your work looks really good. See my notes.*@