`q001. The same process we modeled in class last time has the following characteristics:
The temperatures of the hot and cold sources are T_c and T_h, initial volume is V_0, initial pressure is P_0, water density is rho and the acceleration of gravity is g.
The system is heated at constant volume, starting at T = T_c, until the temperature is T, where T_c < T < T_h, raising water in a thin tube whose end is open to the atmosphere.
The system is allowed to expand at constant pressure until it comes to final temperature T = T_h, displacing water as it expands. The water exits from the end of the tube.
The system is then cooled to its original temperature and pressure, without doing additional work or absorbing more thermal energy from the hot source.
In terms of T_c, T_h, V_0, P_0, rho and g:
How much PE does the system gain in the process?
****
P1= rhogy+P0 --> `d y= (P1-P0)/rhog= (P0(T/Tc)-P0)/rhog
`d V=V0(T/Th)-V0
`d y= (P1-P0)/rhog= (P0(T/Tc)-P0)/rhog∆PE = (rho*g* `d y) `d V= rho*g*(P0(T/Tc)-P0)/rhog(V0(T/Th)-V0)= (P0(T/Tc)-P0)(V0(T/Th)-V0)= V0P0(T/Tc-1)(Th/T-1)
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Assuming the gas to be diatomic, how much thermal energy flows into the system during the process?
****∆Q01= Cv*n* `d T = 5/2R*(V0P0/nRTc)(T-Tc)= 5/2(V0P0/Tc)(T-Tc)
`d Q12= (Cp*n* `d T)+ `d W= 7/2*R(V0P0/nRTc)(T-Tc)= 7/2(V0P0/Tc)(T-Tc)
@& From state 1 to 2 the change in temperature is
`dT_12 - Th - Tc.
This affects your result for the total as well.*@
`d Q= `d Q01+ `d Q12 = 5/2(V0P0/Tc)(T-Tc)+7/2(V0P0/Tc)(T-Tc)= 12/2(V0P0/Tc)(T-Tc)
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What therefore is the ratio of PE gain to thermal energy absorbed by the system?
**** `d PE/ `d Q=(V0P0(T/Tc-1)(Th/T-1))/(12/2(V0P0/Tc)(T-Tc))
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How would your answer differ if the gas was monatomic, and why?
****No rotational Q so 1 less DOF means only 3 DOF thus Q= 1/2 +1/2 +1/2 = 3/2 and for work we add another 2/2 so the total is Q= (3/2+2/2)+(3/2) = 8/2
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`q002. By heating the air in a bottle from 273 K to 300 K, you have raised water in a thin vertical tube to a height of about 100 cm.
By how much will the water level in the tube change, per Kelvin, if the temperature is changed?
****100/(300-273)= 3.704
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Assume the tube has cross-sectional area 0.1 cm^2. This could make a small difference in your preceding answer, which was based on the assumption that the volume of the tube is negligible. We won't worry about that for the preceding, but we will need to consider it for the following.
However the cross-sectional area of the tube makes a difference to the following:
If the tube makes a sharp right-angle turn at the 100 cm height (becoming horizontal), then how many cm of the additional tube will be filled, per Kelvin, if the temperature increases?
****
Not certain about this but here's what I know.
x = hor dist. and T2 is new temp
V1=V0
V2 = V0(T2/300)
Also:
V2= V0+0.1x
x= 10(V2-V0)
x/ `d T=10(V2-V0)/ `d T
x/ `d T= 10(V0(T2/300)-V0)/ `d T
@& Good, but T_2 = (300 + `dT) so I believe your expression will be
x/ `d T= 10(V0 `dT / 300 )/ `d T = 10 V0 / 300.*@
Of course this only considers that the volume in the first phase isn't considered but it did gain 10cm^2 during the first phase.
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Would the our results change if the air in the bottle had initially been at 290 Kelvin (assume that the water level and the position at which the tube makes its right-angle turn are adjusted accordingly)?
****
I'm not certain but I want to say:No, the new change in volume is all that affects the x distance since P=const.
@& I think the result would be 10 V0 / 290.
The effect of increasing the temperature by a degree depends on that one-degree change, as a fraction of the existing temperature. This is because the existing temperature determines the pressure at which the expansion begins.*@
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What would change in our analysis if instead of a right-angle turn, the angle changed by a little less than 90 degrees, so that the tube has an upward slope of 0.1?
**** Here P would also be changing and contributing to the upward movement as well as V change of course both would be related to T change. Seems that the slope of 1/10 means that for every P increase y changes 1 while V will increase x by 10 that of y meaning that `d V has the most significant effect.
@& Good. It might be interesting to express this in terms of differentials.*@
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`q003. A wall is constructed of two different materials of equal thickness, the first with double the thermal conductivity of the other. The inner and outer walls are maintained at a constant temperature difference until a steady flow of thermal energy through the wall is achieved.
At that time, which wall will have the greater temperature gradient, and what will be the ratio of the temperature gradients? Hint: In the steady state, equal amounts of thermal energy pass through both materials.
****Means M2= 2*K1 --> H_2 = 2*H_1
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What therefore is the ratio of the temperature change through the first layer, to the temperature change through the second?
****∆T_M2= 1/2 `d T_M1
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If the inner wall is at temperature 30 Celsius and the outer at temperature 10 Celsius, what is the temperature at the point where the layers meet?
**** It was stated explicitly so I'm taking the inner material to be M_2 from above.
`d T = 20 so 30-2 `d T- `d T=10 --> `d T = 20/3
Thus the middle will be 30-20/3=23.333 º C
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If the first wall is also 50% thick than the other, how will the steady-state temperature gradients compare?
**** This means 1/L2=1/0.5L1 thus T2= (1/0.5)T1= 2T1
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What therefore is the ratio of the temperature change through the first layer, to the temperature change through the second?
**** 1 `d T1=2 `d T2
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If the inner wall is at temperature 30 Celsius and the outer at temperature 10 Celsius, what is the temperature at the point where the layers meet?
****∆T = 20 so 30-4T1- `d T1=10 --> `d T1 = 4 --> T= 26ºC
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`q004. (University Physics) The PE gain of the system analyzed in the previous class is V0 ( (T_h / T_c) * ( P_0 / (P_0 + rho g y) ) - 1) ) * rho g y.
If the system is heated from temperature T_c to temperature T_h, at constant volume, what will be the maximum pressure and what therefore will be the maximum possible value of y?
****P1= P0(Th/Tc)
P1= P0+rhogy --> y_max = P0(Th/Tc-1)/rhog
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Substitute this maximum possible value of y for into the equation and simplify. What is your result, and why does this result make sense?
****PE= V0((T_h / T_c)*(P_0/(P_0 + rho g(P0(Th/Tc-1)/rho g))-1))*rhog(P0(Th/Tc-1)/rho g)
V0((T_h / T_c)(1/(1+Th/Tc-1)-1)(P0(Th/Tc-1)
The ones cancel in the middle which makes that inner expression just 1/(Th/Tc)-1= Tc/Th-1 and that means Th/Tc(Tc/Th-1)= 0
Thus PE= 0
This is somewhat counterintuitive but I know that it is true because I've seen the derivative of the function that relates dPE/dTi where Ti is some intermediate value of T and at Ti=Th--> dPE= 0. Now exactly why this is so I'm not sure but I think it has to do with the fact we're treating `d V= 0 but V does change as the water moves up the tube and if that water can't go up the tube then there's no `d PE.
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If the system is heated from temperature T_c to temperature T_h, at the constant initial pressure P_0, then how much water will be displaced, and to what height y? How much PE change will therefore occur?
**** This very quickly leads to `d PE= 0 since P0=P1--> P0/P1= 1--> ( P_0/(P_0+ rhogy))- 1)) = 0
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We can factor all constant factors from the rest of the expression, obtaining the form
`dPE = rho g V0 * ( (T_h / T_c) * ( P_0 / (P_0 + rho g y) ) - 1) ) * y.
Find the value of y at which this expression is maximized.
****
The derivative wrt to y is: dPE/dy = (rho*g*V0*(T_h/ T_c)) [((-rho*g*P_0 )/ (P_0 + rho g y)^2)*y + (P_0/(P_0 + rho*g*y)-1)]
The max is when dPE/dy= 0. Now I've worked with this and can't find an explicit solution for y the best I came up with is:
y= (P0+rho*g*y)/(rho*g)- (P0+rho*g*y)^2/(P0*rho*g)
When PE is in terms of Ti the maximum is easily found but this doesn't seem to work as well.
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Find the value of `dPE for this maximizing value of y.
**** Because I couldn't get y explicitly this doesn't work out to anything very useful since y is still in the expression
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Sketch a graph of `dPE vs. y.
**** The dPE/dTi I mentioned above is a nice skewed concave down parabola that has it's max at Ti= sqrt (Tc*Th). I think this would have a very similar appearance with the max skewed slightly more toward y0 or 0 cm.
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`q005. The system first analyzed in the preceding class had a rectangular P vs. V graph, due to the way the system was first cooled until it reached the original pressure.
The third phase of the cycle changes if we allow the gas to expand suddenly. Instead of a vertical line segment we get a part of the adiabatic curve P V^gamma = constant. This results in a different initial temperature T_3 and a different initial volume V_3 for the fourth phase of the cycle (that phase is still isobaric, just over a different range of volumes).
What will be the expressions for T_3 and V_3?
**** I'm still a little unsure of the exact way the adiabatic eqs work together but I cobbled together this:
T2V2^(gamma)= T3V3^(gamma-1)
gamma = 7/2 = 3.5
T2=Th
V2=V0*Th/Tc
Th(V0Th/Tc)^2.5 = T3V3^2.5
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The PE change of the system remains the same as before. However the work done by the system does differ from our previous result? What is the expression for the work done by this system?
****W=nCv(Th-T3)
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How does the amount of thermal energy transferred to the cold sink differ between the original cycle and this one?
****
It is greater since T3 is proportional to the 2.5ths power of V and not the 1st power of V. Since V2< V3 --> (V2/V3)^gamma < V2/V3
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`q006. The third phase of the cycle could also be accomplished isothermally, adding thermal energy to the (more slowly) expanding gas until it reaches the original temperature. Since T and n remain constant, the graph will in this case follow an isothermal curve P V = constant. Again T_3 and V_3 will differ from the values obtained for previous cycles, as will the amount of energy absorbed from the hot sink and the amount absorbed by the cold sink.
What will be the expressions for T_3 and V_3?
****T3=Th and V3=V2*P2/P3
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What is the expression for the work done by this system?
****W= Int(P*dV) from V2 to V1 This is an integral because of the nonlinear path the process occurs over, i.e. Int dV ≠ `d V≠ V2-V1 (I think that's why anyway).
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How does the amount of thermal energy transferred to the cold sink differ between the original cycle and this one?
****It is less since one part of the process involves no energy going to the cold sink whereas before 2 phases involved cold sink energy transfer, namely 2-3 & 3-4(0) and now just 3-4(0)
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How does the amount of thermal energy absorbed from the hot sink differ between the original cycle and this one?
****The amount here increases even though the process is isothermal thermal energy is added it just isn't added in the form of heat, it's going into expansion. In other words: phases 0-1 1-2 and 2-3 all involve energy transfer from hot sink.
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@& Good work. I've inserted a couple of notes.
I'll probably post this again shortly, with some additional comments. Let me know if that doesn't appear within about a day.*@