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course Phy 242
February 25 around 5pm.
110214`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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The ratio of pressure is the same as the temperature ratio: (450/300)=(3/2)
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What is the ratio of the highest volume to the lowest?
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The ratio is the same as the temperature ratio: (900/450)=(2)
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In terms of the initial gas volume and pressure V_0 and P_0:
What are the volume and pressure when the system reaches 450 K?
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Volume: V0
Pressure: (3/2)P0
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What are the volume and pressure when the system reaches 900 K?
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Volume: 2V0
Pressure: (3/2)P0
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If the added pressure is used to raise water, how high will it be raised?
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Pressure goes from P0 to (3/2)P0, additional pressure is (1/2)P0 supports water column, water column of height “y” requires pressure (rho g y) at the bottom.
(rho g y) = (1/2)P0
Y=(1/2)(P0/rho g)
=P0/(2 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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Volume goes from V0 to 2V0.
Change is 2V0-V0= V0
So, volume V0 is displaced to height:
Y=(P0/(2 rho g))
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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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V0 * (P0/(2 rho g))
=(P0V0/(2 rho g))
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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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Pressure ratio: (600/300) = 2
Volume ratio: (900/600) = (3/2)
Pressure increase: 2P0-P0 = P0
Volume increase: (3/2)V0 - V0 = (1/2)V0
Height (y) with (rho g y=P0) so y=(P0/(rho g))
(1/2 V0) * (P0/(rho g)) = (V0P0/(2 rho g))
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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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Pressure ratio: (Ti/Tc - 1)
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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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Volume ratio: (Th/Ti - 1)
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What is the expression for the change in the pressure?
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=P0(Ti/Tc) - P0
=P0 (Ti/Tc - 1)
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What is the expression for the change in the volume?
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=V0 (Th/Ti - 1)
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What do we get when we multiply the change in pressure by the change in volume?
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[P0 ((Ti/Tc)- 1)] * [V0 ((Th/Ti) - 1)]
= (P0V0) * ((Ti/Tc)-1)((Th/Ti)-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 very similar.
@& The pressure increases from P_0 to 1.5 P_0, an increase of
`dP = 1.5 P_0 - P_0 = 0.5 P_0.
To support a water column of height y, an addition pressure rho g y is required.
Thus the height y to which water is raised will be given by the equation
rho g y = 0.5 P_0
so that
y = -.5 P_0 / (rho g).
The added gas volume is the difference between the initial volume V_0 and the maximum volume 2 * V_0. The difference is
`dV = 2 V_0 - V_0 = V_0.
The added volume multiplied by the height is therefore
`dV * y = V_0 * (.5 P_0 / (rho g) ) = .5 P_0 V_0 / (rho g).
The weight of the water raised is equal to its mass multiplied the acceleration of gravity. Its mass is equal to its volume multiplied by its density so
m = V_0 * rho
and
weight = m g = V_0 * rho g.
The PE change of the water is equal to the work done in raising it. Thus weight V_0 * rho g is raised through height y = 0.5 P_0 / (rho g), giving us a PE change of
`dPE = weight * height = (V_0 * rho g) * (0.5 P_0 / (rho g) ) = 0.5 P_0 V_0 / (rho g).
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What would we have to do with our expression to get the actual PE gain?
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Divide (P0V0) by (rho*g)
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&#Good responses. See my notes and let me know if you have questions. &#