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course Phy 242
February 25 around 3pm.
110207`q001. The gas in a bottle is heated, forcing water first up a thin tube then out of the top of the tube. After the water reaches its maximum temperature, the top of the tube remains open to the atmosphere, the heat source is removed and the gas is allowed to cool until the water in the tube has nearly reached the level of the water in the bottle. A valve is closed, preventing any air from entering the system through the tube, and the system is allowed to continue cooling to room temperature.
Sketch a pressure vs. volume graph for this cycle.
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OK
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What shape is made on your graph?
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A square/box
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`q002. Suppose that in the preceding we have a large supply of cool air at temperature T_c and a large supply of warm water at temperature T_h. The gas in the bottle begins at temperature T_0 = T_c and pressure P_0, with volume V_0,the tube's open end is at height y relative to the water in the bottle. We will raise the temperature of the gas in the bottle to temperature T_h by bathing the bottle in warm water, and we will allow it to cool while surrounded by cool air.
In terms of T_c, P_0, V_0 and y, as well as the density rho of water and the acceleration g of gravity (not all of which will necessarily appear in every expression), what will be the pressure P_1, volume V_1 and temperature T_1 when water first reaches the top of the tube?
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P1 = P0 + (rho g y)
V1 = V0
T1 = T0(P1/P0) or Tc((rho g y+P0)/P0)
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In terms of the same variables, what will be the pressure P_2, volume V_2 and temperature T_2 when the gas reaches its maximum temperature T_h?
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P2 = P1 = (rho g y + P0)
V2 = V1(T2/T1) or V0 (Th/Tc)(P0/(rho g y + P0))
T2 = Th
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In terms of the same variables, what will be the pressure P_3, volume V_3 and temperature T_3 when the water has in the tube has receded to just above the level of the water in the bottle and the tube is closed?
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P3 = P0
V3 = V1(T2/T1) or V0 (Th/Tc)(P0/(rho g y + P0))
T3 = T2(P3/P2) or Th(P0/(rho g y + P0))
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In terms of the same variables, what will be the pressure P_4, volume V_4 and temperature T_4 when the system has cooled to its original temperature T_c?
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P4 = back to state 0
V4 = back to state 0
T4 = Tc
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`q003. In the preceding:
How much thermal energy enters or leaves the system during each of the four separate phases of the cycle? Assume the gas to be diatomic.
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1/2nR’dT per degree of freedom
2/2nR’dT for volume change at constant pressure
State 0 to 1: (5/2nR’dT)
State 1 to 2: (5/2nR’dT + 2/2nR’dT)
State 2 to 3: (5/2nR(T3-T2))
State 3 to 4: (7/2nR(T4-T3))
@& T0, T1, T2, T3 and T4 need to be expressed in terms of the given variables.*@
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How much thermal energy is taken from the warm water?
@& From State 0 through State 2 the system is being heated, and the energy required for the heating is taken from the warm water.*@
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I’m not sure what this or what the next one would be…
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How much thermal energy is added to the cool air?
@& From state 2 through state 4 the system is cooling, with energy being taken from the system by the cool water.*@
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How much potential energy is gained by the water which flows out of the top of the tube, assuming it is caught in a reservoir at that height?
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‘dPE = (V2-V1) * (rho g y)
=[(V0*(Th/Tc)*(P0/(rho g y + P0))-V0)] * (rho g y)
@& The details:
The temperature changes for the four phases are found using the temperatures found in the preceding.
During the first and third phases the temperature changes at constant volume, so the five degrees of freedom imply thermal energy required 5/2 n R `dT.
During the second and fourth phases work is done against pressure (positive work in phase 2, negative work in phase 4) and the thermal energy required is 7/2 n R `dT.
Putting this information together with the expressions for temperature changes we obtain the expressions for the thermal energy transfers.
During the first and second phases thermal energy is taken into the system from the warm water, and during the third and fourth phases thermal energy is taken from the system by the cool water.
The expressions for these quantities are found in the Class Notes.
The water that flows out of the top of the tube has volume equal to the change in volume during phase 2:
`dV = V_0 * (T_h / T_c) * P_0 / (P_0 + rho g y) - V_0
and hence mass
mass = rho `dV = rho ( V_0 * (T_h / T_c) * P_0 / (P_0 + rho g y) - V_0 ).
The weight of this water is
weight = mass * g = rho * g ( V_0 * (T_h / T_c) * P_0 / (P_0 + rho g y) - V_0 )
so its gravitational PE increases by
`dPE = weight * change in vertical position
= ( rho * g ( V_0 * (T_h / T_c) * P_0 / (P_0 + rho g y) - V_0 ) ) * y.
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`q004. We are still referring to the parameters of the system in question 2.
How high would it be possible to raise water in the tube by heating the system from temperature T_c to temperature T_h, assuming the tube to be sufficiently long?
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Change in Q:
(5/2 nR (Tc(rho g y + P0/P0) - Tc) + (7/2nR(Th-Tc(rho g y+P0/rho g y)))
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Let's call this height y_max.
If the height y is half as great at y_max, then what is expression for the PE change of the water?
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PE change = (volume of water raised)*(density)*(accel of grav)*(change in vertical position)
=(V2-V1)*(rho g y)
=(V0 (Th/Tc)(P0/rho g y + P0)-V0) * (rho g y)
If the height is half as great at y_max, wouldn’t it be “y_optimal” on the graph?
@& The optimal height is not half of y_max. Close, but not the same.*@
@& If the function was quadratic in y then the graph of PE vs. y would be parabolic and the halfway point would optimize.*@
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If the height y is adjusted to be a little greater than half of y_max, will the PE change increase or decrease relative to your preceding answer? University physics students could answer this question by considering the derivative of an appropriate function.
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Derivative of (change in PE/change in Q_in)= (use product, quotient and chain rule)
= [(V0 (Th/Tc)(P0/rho g y + P0)-V0) * (rho g y)] / [(5/2 nR (Tc(rho g y + P0/P0) - Tc) + (7/2nR(Th-Tc(rho g y+P0/rho g y)))]
Taking the derivative:
= [(V0*rho g y((Th/Tc)(P0/P0+rho g y))-1)] / [1/2nR(5 Tc(rho g y+P0/P0)-1) + 7(Th-Tc(rho g y+P0/P0))]
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If y = r * y_max, what is the expression for the PE change of the water?
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(rho*g*amount of water raised)*(height of water that’s raised or (r*y_max))
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For what value of r will the PE change of the water be maximized? (University Physics students only)
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Y_optimal on the graph
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For what value of r will the ratio of PE change to energy absorbed from the warm water be maximized? (University Physics students only)
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If y=r*y_max then dPE and (dPE/dQ_in) become functions of “r”?!
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@& You need to find the expression for y_max in terms of the given variables. Then subsequent expressions will use this value of y_max.
y_max is a fixed quantity for the given temperatures. There is no variable involved in the expression for the PE change when y is, say, the halfway height.
Then if y = r * y_max, y again becomes variable, and the PE becomes a function of y, which is in turn a function of r (a chain rule situation).
If the system is kept at constant volume, as will be the case if the tube is thin, the pressure will rise to
P_1 = P_0 * T_h / T_c,
a change of
`dP = P_0 * T_h / T_c - P_0 = P_0 ( T_h / T_c - 1).
This pressure change will support a column of water of height y_max, such that
rho g y_max = P_0 ( T_h / T_c - 1)
so that
y_max = P_0 ( T_h / T_c - 1) / (rho g).
If the height y is half as great at y_max, then what is expression for the PE change of the water?
Half of y_max is y_max = 1/2 P_0 ( T_h / T_c - 1) / (rho g) = P_0 ( T_h / T_c - 1) / (2 rho g).
Change in gravitational PE when the water is raised to height y was found in a preceding problem to be
`dPE = ( rho * g ( V_0 * (T_h / T_c) * P_0 / (P_0 + rho g y) - V_0 ) ) * y
so when y = y_max we have
`dPE = ( rho * g ( V_0 * (T_h / T_c) * P_0 / (P_0 + rho g y) - V_0 ) ) * P_0 ( T_h / T_c - 1) / (2 rho g)
= 1/2 P_0 V_0 * ( (T_h / T_c) * P_0 / (P_0 + rho g y) - 1 ) ) * ( T_h / T_c - 1)
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`q005. (University Physics students only). If after reaching State 2, instead of letting the water gradually recede as the gas cools, we open a vent in the bottle and simply let the pressurized air suddenly escape, the pressure and volume will change in such a way that P V^gamma = constant, where gamma = C_p / C_v (gamma = 7/2 for a diatomic gas, 5/2 for a monatomic gas).
In this case what will be the temperature T_3 and the volume V_3 when the pressure has reached the original pressure P_3 = P_0?
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=(P2V2/T2) = (P3V3/T3)
=P2V2 - P3V3
@& Good start, but so far you have expressed only one equation, and you need to find two quantities, P3 and T3.
The expansion is adiabatic, so that
P V ^ gamma = constant,
where gamma = C_p / C_v = (7/2 R) / (5/2 R) = 1.4.
You have to combine this with the fact that
(P2V2/T2) = (P3V3/T3)
in order to solve for the two quantities T_3 and V_3.
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How much work does the gas do in this expansion?
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How would I solve for this?
@& As you have seen previously, the work done in an expansion is the area beneath the P vs. V graph, i.e., the integral of P with respect to V..
In this case P V ^ gamma = constant, so
P = constant * V^(-gamma).
You know P2 and V2, so you can find the value of the constant by simply calculating P2 * V2^gamma (you could just as well use P3 and V3).
Having done so you can integrate P with respect to V, between V2 and V3.
You should do this and compare your expression to the area under the curve between state 1 and state 2.
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If the gas is then permitted to cool at constant pressure until it has reached its original temperature T_c, how much thermal energy must be transferred from the gas to the surrounding air?
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You know that P3 V3 = P2 V2, so V3 = V2 * (P2 / P3), where P3 is just P0.
In this case the temperature remains constant, so P V = constant.
You can evaluate the constant by calculating P2 * V2, or if you prefer P3 * V3.
Thus you will have the expression for P:
since P V = constant and you have an expression for the constant,
P = constant / V.
Expressed in terms of the original parameters, this is a function of V that can be integrated with respect to V from V = V2 to V = V3.*@