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course Phy 242
2-27
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?
Pressure ratio = 450K/300K = 3/2,
because pressure builds from T_1 to T_2, where T_1 = 300K and T_2 = 450
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What is the ratio of the highest volume to the lowest?
Volume ratio = 900K/450K = 2,
because volume expands from V_1 to V_2, where V_1 = 450K and V_2 = 900
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In terms of the initial gas volume and pressure V_0 and P_0:
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What are the volume and pressure when the system reaches 450 K?
Pressure = (3/2)P_0
Volume = V_0
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What are the volume and pressure when the system reaches 900 K?
Pressure = (3/2)P_0
Volume = 2V_0
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If the added pressure is used to raise water, how high will it be raised?
The added pressure = (1/2)P_0
So I believe the question is inferring that the (1/2)P_0 is used to support water column
For rho*g*y = (1/2)P_0
y = P_0/2rho*g
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If the added gas volume displaces water to the preceding height, what volume of water will be raised?
`dV = 2V_0 - V_0 = V_0
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What do we get if we multiply the volume of water raised by the height to which it is raised?
y*V_0
y*V_0
= (P_0/(2rho*g))*V_0 =( P_0*V_0)/(2rho*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.
Pressure ratio = 600K/300K = 2,
because pressure builds from T_1 to T_2, where T_1 = 300K and T_2 = 600K
Volume ratio = 900K/600K =(3/2),
because volume expands from V_1 to V_2, where V_1 = 600K and V_2 = 900
At Temp of 600K
Pressure =2P_0
Volume = V_0
At Temp of 900K
Pressure =2P_0
Volume = (3/2)V_0
The added pressure = 2P_0 - P_0 = P_0
For rho*g*y = P_0
y = P_0/rho*g
`dV = (3/2)V_0 - V_0 = (1/2)V_0
`dV*y =
y*(1/2)V_0
= (P_0/(rho*g))*(1/2)V_0 =( P_0*V_0)/(2rho*g)
Same final expression as previously.
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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?
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?
V_0(T_h/T_i)
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What is the expression for the change in the pressure?
P_0(T_i/T_c) - P_0 = P_0( (T_i/T_c) - 1) = `dP_0
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What is the expression for the change in the volume?
V_0(T_h/T_i) - V_0 = V_0( (T_h/T_i) - 1) = `dV_0
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What do we get when we multiply the change in pressure by the change in volume?
P_0( (T_i/T_c) - 1) *V_0( (T_h/T_i) - 1)
=P_0*V_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?
PE_gain = volume of water*ht raised*g
We have volume of water and from pressure term it would not be hard to find ht to which water is raised.
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What would we have to do with our expression to get the actual PE gain?
For rho*g*y = P_0( (T_i/T_c) - 1)
y = P_0( (T_i/T_c) - 1) /rho*g
So subbing into our equation of P_0( (T_i/T_c) - 1) *V_0( (T_h/T_i) - 1)
We let pressure term = P_0( (T_i/T_c) - 1) /rho*g
PE_gain = volume of water*ht raised*g
V_0( (T_h/T_i) - 1) * P_0/rho*g( (T_i/T_c) - 1) *g
Which is
= Volume of water raised*ht raised to*accel of gravity
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02-27-2011