course Phy: 202 Ic_class_100208The following results from the kinetic theory of gases are well confirmed by various experiments:
100208
The following results from the kinetic theory of gases are well confirmed by various experiments:
We can write these quantities as follows:
`dQ = C_v * n * `dT, where C_v = 3/2 R for a monatomic gas and C_v = 5/2 R for a diatomic gas.
We call C_v the specific heat at constant volume.
`dQ = C_p * N * `dT, where C_p = 5/2 R for a monatomic gas and C_p = 7/2 R for a diatomic gas.
We call C_p the specific heat at constant pressure.
Using these facts, the ideal gas law PV = n R T, with atmospheric pressure 100 000 Pa or 100 kPa ('kPa' stands for 'kilopascal', or 1000 Pa, so that 100 kPa = 100 000 Pa), we analyze a 'bottle engine' consisting of a bottle containing some water and some air, with a vertical tube whose low end is below water level and whose upper end extends to an unspecified height above the level of the water. The tube is considered 'thin', so that its volume is negligible compared to that of the gas in the bottle.
If the gas in the bottle is heated, the increased pressure will 'push' water up the tube until it reaches the end of the tube. Since the tube is 'thin' the volume of water displaced will be negligible, so the volume of gas in the system will change by only a negligible amount. The volume of the gas will therefore be essentially constant during this process.
If the system continues heating after water is raised to the end of the vertical tube, water will begin flowing out of the tube and the gas in the system will be free to expand by displacing water. The displaced water ends up at the height of the end of the tube, so its gravitational PE is increased. The mechanical energy of the system is therefore increased.
Assuming that we are using a monatomic or a diatomic gas:
We can also calculate the new pressure.
We can also calculate the new volume, and hence the change in volume.
We are going to develop a symbolic analysis of a system as it is heated. You will then be asked a series of questions.
Specifically, let's assume initial temperature, pressure and gas volume to be T0, P0 = P_atm and V0. Let T1 be the temperature to which the system is heated at constant volume, and T2 the temperature to which the system is heated at constant pressure. We consider the following questions:
Since P V = n R T, we have n = P V / (R T) so that in the first state,
n = P0 V0 / (R T0).
The gas in the system is closed to the atmosphere so this will be the value of n throughout the process.
Answer: We will first find the pressure, the the height of the vertical water column supported by the excess pressure.
Volume remains constant, as does the number n of moles of gas. It follows that P / T = n R / V remains constant and
P1 / T1 = P0 / T0 so that
- P1 = P0 * (T1 / T0).
The pressure at the top of the water column is P0 and the pressure at the bottom is P1. The pressure difference between the top of the column and the bottom is therefore P1 - P0.
The pressure difference is also equal to rho g h, where h is the height of the column.
Thus rho g h = P1 - P0 and
h = (P1 - P0) / (rho g).
You will be asked below to express this in terms of V0, T0, P0 and T1.
Answer: This occurs at constant volume. The internal energy therefore changes by
`dU_01 = C_v * `dT = C_v * (T1 - T0),
where C_v = 3/2 R if the gas is monatomic and C_v = 5/2 R if the gas is diatomic.
We have used `dU_01, with the subscript 01, to denote the change in internal energy from State 0 (at V0, T0, P0) to State 1 (at V1, T1, P1).
Answer: During this phase the pressure remains constant, since the water column doesn't get any higher. Thus P and n remain constant, so V / T remains constant.
Thus V2 / T2 = V1 / T1 and V2 = V1 * (T2 / T1).
Since volume didn't change during the first phase, V1 = V0 and we conclude that
V2 = V0 * (T2 / T1).
Answer: We just need to find the difference between V1 and V2.
The volume of the gas changed from V1 = V0 to V2 = V0 * (T2 / T1). So the change is
`dV_gas = V2 - V1 = V0 * (T2 / T1) - V0 = V0 * ( T2 / T1 - 1)
The only way the volume of the gas can change is to displace an equal amount of water. So the amount of water flowing out of the tube is
`dV_water = `dV_gas = V0 * ( T2 / T1 - 1).
Answer: The system was at constant pressure so
`dU_12 = C_p * `dT = C_p * (T2 - T1),
where C_p = 3/2 R if the gas is monatomic and C_p = 5/2 R if the gas is diatomic.
We have used `dU_12, with the subscript 12, to denote the change in internal energy from State 1 (at V1, T1, P1) to State 2 (at V2, T2, P2).
Answer: The only part of the system that changes its vertical position (other than the negligible amount of water which fills the tube) is the water that flows out.
We have found that volume `dV_water = V0 * (T2 / T1 - 1) flows out at height h = (P1 - P0) / (rho g).
The mass of volume `dV_water is just the product of the volume the density of water, mass = rho * `dV_water.
The weight of the water is therefore the product of its mass and the acceleration of gravity, weight = mass * g = rho * g * `dV_water.
The work required to raise the water is weight * change in altitude = weight * h, or
- `dW = (rho * g * `dV_water) * (P1 - P0) / (rho g) = `dV_water * (P1 - P0).
`q001. Water is to be raised to the top of a this open tube, at a distance of 1.3 meters above the level of the bottle. The initial pressure of the water in the bottle is 1 atmosphere, about 100 kPa or 100 000 Pa, and the initial temperature is 273 K. The initial gas volume is 2.5 liters, and the gas is diatomic.
How much additional pressure will be required?
****
Plain old rho g y would work for this, but let's put it in the context of Bernoulli's Equation:
Let point A be at the top of the water column in the tube, point B at water level in the bottle.
The pressure at the top of the tube will be atmospheric pressure, so P_A = 1 atm = 100 000 Pa = 1 kPa.
Let pressure at the level of water in the bottle be designated P_B. This is the pressure we want to find.
If we let the water level in the bottle be y = 0, then y_B = 0 and y_A = 1.3 m.
Presumably nothing moves fast enough to make 1/2 rho v^2 significant, so we will assume that v_A = v_B = 0.
From Bernoulli's Equation we get
1/2 rho v^2 + rho g y + P = constant.
Leaving off 1/2 rho v^2, which is zero at both points we conclude that
rho g y_B + P_B = rho g y_A + P_A.
Solving for P_B we get
P_B = rho g y_A - rho g y_B + P_A,
and substituting our know values we have
P_B = 1000 kg/m^3 * 9.8 m/s^2 * 1.3 m - 1000 kg/m^3 * 9.8 m/s^2 * 0 m + 100 000 Pa = 113 000 Pa, approx..
This is 13 000 Pa higher than the original pressure in the bottle.
#$&*
To what temperature must the gas be heated to achieve the necessary pressure?
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n and V are considered constant, so P / T will be constant
P0 / T0 = P1 / T1 so
T1 = T0 * (P1 / P0).
P0 is atmospheric pressure 100 000 Pa, or 100 kPa and P1 is 113 000 Pa, or 113 kPa.
T0 = 273 K.
So
T1 = 273 K * ( 113 kPa / (100 kPA) ) = 310 K, approx..
#$&*
How many moles of gas are present, and how many Joules of energy must therefore be transferred to the bottle in order to heat the air to the necessary pressure?
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n = P V / (R T). Using the original pressure, volume and temperature we have
n = 100 000 Pa * (.0025 m^3) / (8.31 J / (mol K) * 273 K) = .11 moles, approx..
The gas is said to be diatomic, and the expansion is at constant volume, so the change in the internal energy of the gas is
`dU = 5/2 n R * `dT.
`dT is (very approximately) 40 K so
`dU = 5/2 * .11 mol * 8.31 J / (mol K) * 40 K = 90 J, again very approximately.
The expansion is at constant volume so no work needs to be done against the pressure, so thermal energy necessary is equal to this change in internal energy
`dQ = `dU = 90 J, approx..
#$&*
`q002. This question continues the preceding. If after raising water to the specified level, the gas is then heated to 373 K at constant pressure, then
What will be its new volume?
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n and P are constant in this process, so P2 / T2 = P1 / T1, and therefore
V2 = V1 * (T2 / T1) = 2.5 liters * (373 K / 310 K) = 3 liters, approx.
#$&*
How much water must be displaced to accommodate the new volume?
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The gas must expand from 2.5 liters to about 3 liters, an expansion of about .5 liters.
In order for the gas in the bottle to expand, then, .5 liters of water must be displaced
#$&*
If the displaced water is caught in a reservoir at the level of the top of the tube, by how much does its gravitational PE increase?
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The top of the tube is 1.3 m higher than the water in the bottle.
.5 liters of water has mass .5 kg and therefore weight 4.9 N.
So the PE of the water increases by
`dPE = 4.9 N * 1.3 m = 6.4 Joules, approx..
#$&*
How much thermal energy is required to heat the water from its previous temperature to 373 K?
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The temperature changes from 310 K to 373 K, a temperature change we will approximate here as 60 K.
The gas is diatomic and is expanding at constant pressure, so C_p = 7/2 R and
`dQ = C_p `dT = 7/2 n R `dT = 7/2 * 8.31 J / (mol K) * .11 mol * 60 K = 200 J, very approximately.
#$&*
The next two questions are essentially identical to the first two questions, but you are asked to use symbols to represent your solutions:
`q003. In terms of the symbols P0, V0, T0, and h, as well as R for the gas constant:
What is the expression for the temperature at which water is raised to height h in the tube?
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To raise water to height h, we can use either Bernoulli's Equation or `dP = rho g `dy to see that gas pressure must change by
`dP = rho g h.
As observed in the solution to the first question in this set,
T1 = T0 * (P1 / P0).
Now P1 = P0 + `dP = P0 + rho g h, so
T1 = T0 * ( P0 + rho g h ) / P0
#$&*
What is the expression for the number of moles of gas in the system?
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n = P V / (R T), so using P0, V0 and T0 we obtain
n = P0 V0 / (R T0).
#$&*
If you know the change in temperature and the number of moles how do you find the thermal energy required to achieve the temperature change?
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`dQ = C * n * `dT, where C = C_v if the temperature changes at constant volume, and C = C_p if at constant pressure.
#$&*
What is the expression for the change in temperature?
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We obtained the expression T1 = T0 * ( P0 + rho g h ) / P0, so
`dT = T1 - T0 = T0 * ( P0 + rho g h ) / P0 - T0.
This expression can be simplified a bit, but to avoid confusion about the algebra we'll leave it unsimplified for now.
#$&*
What is the expression for the thermal energy required?
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This is a diatomic gas at constant volume, so C = C_v = 5/2 R. The thermal energy required is
`dQ_01 = 5/2 R * n * `dT,
where the subscript _01 denotes the change from state 0 to state 1.
n = P0 V0 / (R T0) and `dT = T1 - T0 = T0 * ( P0 + rho g h ) / P0 - T0. The result is sorta ugly but it isn't hard to write down:
`dQ_01 = 5/2 R * [ P0 V0 / (R T0) ] * [ T0 * ( P0 + rho g h ) / P0 - T0 ],
where the brackets [ ] have been used to 'set off' the expressions for n and `dT.
There are two simplifications at this point. The first is obvious; we have R in a numerator of one factor, and R in the denominator of another, so R 'divides out' and we have
`dQ_01 = 5/2 * [ P0 V0 / T0 ] * [ T0 * ( P0 + rho g h ) / P0 - T0 ].
The second is a little less obvious, but we have T0 in the denominator of one term and T0 as a factor of each term in the numerator of another. We get
`dQ_01 = 5/2 * [ P0 V0 ] * [ T0 * ( P0 + rho g h ) / P0 - T0 ] / T0.
Distributing the division through the last term we have
`dQ_01 = 5/2 * [ P0 V0 ] * [ (T0 * ( P0 + rho g h ) / P0) / T0 - T0 / T0],
which simplifies to
`dQ_01 = 5/2 * [ P0 V0 ] * [ ( P0 + rho g h ) / P0 - 1 ].
#$&*
What are the symbolic expressions for P1, T1 and V1, the pressure, temperature and volume in the resulting state, which we will call state 1?
First, since the volume remains constant, we have V1 = V0.
Form above we have
which in addition to
specifies the state of the system in state 1.
`q004. We now consider the change from state 1 (the state reached in the previous problem) to state 2, at which the temperature is T2. Water in the tube doesn't go any higher, so the pressure remains constant. The temperature increases while the gas remains at the pressure it attained in the preceding.
In terms of the symbols P0, V0, T0, T2 and h, as well as R for the gas constant:
What is the expression for the volume of the gas when the temperature reaches T2?
****
We begin in state 1, where
and the temperature rises to T2 while pressure remains constant.
In this process n and P remain constant so V / T is constant. We conclude that
V2 / V1 = T2 / T1 so that
V2 = V1 * ( T2 / T1).
We must express these quantities in terms of R, P0, V0, T0, T2 and h. Substituting T1 = T0 * ( P0 + rho g h ) / P0 and V1 = V0 into V2 = V1 * ( T2 / T1) we get
V2 = V0 * T2 / ( T0 * ( P0 + rho g h ) / P0 ) = V0 * T2 * P0 / (T0 * ( P0 + rho g h ) ).
We can keep the expression in this form. However we can write the expression in a form that isn't too difficult to make sense of. The following form clearly shows the important ratios T2 / T0, the ratio of final to initial temperature, and P0 / (P0 + rho g h), the ratio of initial pressure to the pressure required to support the water column:
#$&*
What is the expression for the change in the volume of the gas?
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The change in the volume of the gas is
`dV = V2 - V1 = V0 * ( T2 / T0) * (P0 / (P0 + rho g h) ) - V0
We can factor out V0 to get
#$&*
What is the expression for the volume of water displaced?
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The expanding gas displaces water. The volume of water displaced is therefore equal to the change `dV in the volume of the gas:
The quantity (T2 / T0) * (P0 / (P0 + rho g h) ) - 1 is just the ratio by which the volume of the gas increases in the expansion. We can make sense of this.
T2 / T0 is the ratio of highest to lowest temperature and is the same as the ratio by which volume would change if pressure remained constant throughout the process. The volume changes by a factor less than this ratio because in our first phase the gas was heated at constant volume.
P0 / (P0 + rho g h) is the ratio of initial pressure to state 1 pressure, state 1 pressure being the pressure required to support the water column. The latter pressure is the greater so this ratio is less than 1. The greater the height h of the air column, the smaller this ratio and the less the ultimate volume change.
If we just let the gas expand at constant pressure, without raising the water level in the tube, then our final volume would just be V0 * (T2 / T0).
If we first increase the pressure to raise the water level, then we get final volume V0 * (T2 / T0) * (P0 / (P0 + rho g h) ), less than the V0 * (T2 / T0) we would get if we didn't raise the pressure.
Our final expression V0 * ( (T2 / T0) * (P0 / (P0 + rho g h) ) - 1) for `dV represents simply the difference between the initial volume V0 and the final volume V0 * (T2 / T0) * (P0 / (P0 + rho g h) ) .
Our expression for the final volume V2 could be written
- state 2 volume = state 0 volume * temperature ratio highest to lowest * pressure ratio initial to state 1
and our expression for `dV as
- volume change = state 0 volume * ( (temperature ratio highest to lowest * pressure ratio initial to state 1) - 1),
where the -1 has the effect of simply subtracting the initial volume from the final volume.
... ratio of temperature increase / ratio of pressure increase ...
... percent increase in temperature / percent increase in pressure ...
#$&*
What is the expression for the change in the gravitational PE of the water?
****
A volume `dV_water is raised vertically through displacement h.
The mass of the water is
mass_water = rho_water * `dV_water,
its weight is the product of its mass and the acceleration of gravity
weight_water = mass_water * g = rho_water * g * `dV_water
and the change in its gravitational PE is
`dPE = weight_water * vertical displacement = rho_water * g * `dV_water * h
To put this in terms of the given quantities we need only replace `dV_water with the expression derived previously:
`dPE = rho_water * g * V0 * ( (T2 / T0) * (P0 / (P0 + rho g h) ) - 1) * h, which we rearrange slightly to get
#$&*
What is the expression for the amount of thermal energy required?
****
The thermal energy required is
`dQ = C * n * `dT,
where in this case C is the molar specific heat of a diatomic gas at constant pressure, C_p = 7/2 R.
The temperature goes form T1 = T0 * ( P0 + rho g h ) / P0 to T2 so `dT = T2 - T0 * ( P0 + rho g h ) / P0.
The number of moles is n = P0 V0 / (R T0).
So the thermal energy required is
`dQ_12 = (7/2 R) * (P0 V0 / (R T0) * (T2 - T0 * ( P0 + rho g h ) / P0),
where the subscript _02 denotes the change from state 0 to state 1.
We have R in numerator and denominator so this term will divide out, giving us
`dQ_12 = (7/2) * (P0 V0 / T0 * (T2 - T0 * ( P0 + rho g h ) / P0).
We can further simplify this by distributing the division by T0 over the expression (T2 - T0 * ( P0 + rho g h) / P0):
`dQ_12 = (7/2) * (P0 V0 * (T2 / T0 - T0 / T0 * ( P0 + rho g h )/ P0), which simplifies to
To understand this final result for `dQ_12, note the following:
We could write `dQ_12 in the following form:
`dQ_12 = 7/2 P0 V0 * ( temperature ratio highest to lowest - pressure ratio state 1 to state 0)
Our expression for `dPE is
`dPE = (rho_water * g * h) * V0 * ( (T2 / T0) * (P0 / (P0 + rho g h) ) - 1)
#$&*
`q005. Sketch a graph of pressure vs. volume, from state 0 to state 1 to state 2,
A gas is heated at constant volume from state 0, where pressure is 100 000 Pa and volume is 1.5 liters, to state 1, where pressure is 150 000 Pa. It then expands at constant pressure to state 2, where the volume is 2.0 liters.
What are the pressure and volume in State 0?
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P0 = 100 000 Pa and V0 = 1.5 liters, which can also be expressed as .0015 m^3.
#$&*
What are the pressure and volume in State 1?
****
The gas is heated at constant volume so the volume is unchanged.
We have
P1 = 150 000 Pa and V1 = V0 = 1.5 liters or .0015 m^3.
#$&*
What are the pressure and volume in State 2?
****
The expansion from state 1 to state 2 is at constant pressure, so
P2 = P1 = 150 000 Pa and
V2 = 2.0 liters or .002 m^3.
#$&*
Each state is represented by a point on a graph of pressure vs. volume. State 0, for example is represented by the point (1.5 liters, 100 000 Pa). What are the coordinates of the points representing State 1 and State 2?
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State 1 is represented by the point (1.5 liters, 150 000 Pa), and state 2 by the point (2.0 liters, 150 000 Pa).
#$&*
Sketch a graph of P vs. V and plot your three points. Sketch a line segment from the first point to the second. Is this segment horizontal, vertical or at a nonzero angle with both vertical and horizontal?
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The points are plotted on a graph with P as the 'vertical' coordinate and V as the 'horizontal' coordinate, consistent with the standard orientation of a graph of y vs. x.
Both state 0 and state 1 share the same volume 1.5 liters, so the state 1 point will lie vertically above the state 0 point. The line segment connecting these points is therefore vertical.
#$*&
Sketch a line segment from the second point to the third. Is this segment horizontal, vertical or at a nonzero angle with both vertical and horizontal?
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Both state 1 and state 2 share the same pressure 150 000 Pa, so the state 2 point will lie horizontally to the right of the state 1 point. The line segment connecting these points is therefore horizontal.
#$&*
On a graph of P vs. V, what quantity stays constant on a vertical line segment?
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P is the 'vertical' coordinate, so P changes as you move up or down.
V is the 'horizontal' coordinate, which doesn't change when you move vertically up or down on the graph.
So V is constant on a vertical line segment.
#$&*
On a graph of P vs. V, what quantity stays constant on a horizontal line segment?
****
V is the 'horizontal' coordinate, which doesn't change when you move vertically up or down on the graph.
P is the 'vertical' coordinate, so P changes as you move up or down, does not change as you move horizontally right or left.
So P is constant on a horizontal line segment.
#$&*
For a constant-volume process, should the line be horizontal, vertical, or at a nonzero angle with both vertical and horizontal?
****
V is the 'horizontal' coordinate, which doesn't change when you move vertically up or down on the graph.
So a constant-volume process is represented by a vertical line segment.
#$*&
For a constant-pressure process, should the line be horizontal, vertical, or at a nonzero angle with both vertical and horizontal?
****
P is the 'vertical' coordinate, so P changes as you move up or down, does not change as you move horizontally right or left.
So a constant-pressure process is represented by a horizontal line segment.
#$&*
`q006. In the notes given prior to the first problem, recall that the volume of water displaced as temperature goes from T1 to T2 is `dV_water = V0 * (T2 / T1 - 1), the PE gain of the system is `dPE = `dV_water * (P1 - P0), and P1 = T1 / T0 * P0.
For a system in which the temperature goes from T0 = 300 K to T1 = 360 K, with the gas remaining at constant volume, what is the value of the ratio T1 / T0? What therefore is the ratio P1 / P0 of the pressures?
****
T1 / T0 = 360 K / (300 K) = 1.2
Intuitively you should know that when volume is constant, pressure goes up with temperature, so you should surmise that
P1 / P0 = T1 / T0 = 1.2.
This can be confirmed easily enough, reasoning from P V = n R T:
Since P / T is constant when n and V are constant,
P0 / T0 = P1 / T1. This is easily rearranged (just multiply both sides by T1 / P0) to obtain the form
T1 / T0 = P1 / P0
#$&*
What would happen to the air column of a 'pressure tube', as the system goes from state 0 to state 1? Sketch the air columns for the two pressures, and give a good description of your sketch.
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The pressure increases from 1 atmosphere to 1.2 atmospheres.
When n and T are constant, which is the very nearly the case for the air column in the pressure tube, the product P * V is constant.
So when P increases, V decreases. P and V are inversely proportional to one another, so
V1 / V0 = P0 / P1 = 1 atm / (1.2 atm) = .83,
and we conclude that
V1 = .83 V0.
Since the tube has constant cross-sectional area, its length must therefore decrease to .83 of its original length.
Your picture should depict an air column that gets shorter by about 1/6 of its original length.
#$&*
For a system in which the temperature goes from T1 = 360 K to T2 = 480 K, with the gas pressure remaining constant, what is the value of the ratio T2 / T1? What therefore is the value of T2 / T1 - 1? Assuming the previous initial volume of 1.5 liters, what is the volume of the water displaced, as a percent of the original water volume?
****
At constant pressure, V2 / V1 = T2 / T1 = 480 / 360 = 4/3 = 1.33 approx.
T2 / T1 - 1 = 4/3 - 1 = 1/3 (or using the approximation T2 / T1 - 1 = 1.33 - 1 = .33).
The gas expands by a factor if 1/3 or about 33%.
The volume of water displaced is equal to the increase in the volume of the air, so the volume of the displaced water is 1/3, or about 33%, of the original volume.
This wasn't requested, but we observe that the original 1.5 liter volume of the gas therefore increases by a factor of 1/3, adding volume `dV = 1/3 * 1.5 liters = .5 liters and ending up with volume 2.0 liters, in the process displacing .5 liters of water.
#$&*
Sketch the bottle and the water in it, indicating how the water level will change as the system goes from state 1 to state 2. Give a good description of your picture.
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Your picture of the original state should depict a bottle with 1.5 liters of air and at least .5 liters of water.
The air volume increases by .5 liters, so the air volume in state 2 should be 1/3 greater than in state 1, or 1.33 times the state 1 air volume. The water volume will have to decrease by .5 liters. Your picture should reasonably depict this.
#$&*
Suppose the system is returned to its original state, then again heated from temperature T0 = 300 K to T2 = 480 K. However unlike before, when state 1 occurred at temperature 360 K, the temperature in state 1 is 420 K. What difference will this make in your answers to the questions you just answered?
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The pressure ratio from state 0 to state 1 will be
P1 / P0 = T1 / T0 = 420 K / (300 K) = 1.4.
The volume ratio from state 1 to state 2 will be
V2 / V1 = T2 / T1 = 480 K / (420 K) = 1.14, approx.
The new water volume is therefore 1.14 * V1 = 1.14 * V0 (recall that V1 = V0 because the first phase occurred at constant volume).
The change in water volume will therefore be
V2 - V1 = 1.14 V0 - V0 = .14 V0.
If the original gas volume was V0 = 1.5 liters, the change in volume is .14 * 1.5 liters = .21 liters.
#$&*
Substitute the expression for `dV_water, and also the expression for P1, into the expression for `dPE. What is the resulting expression for `dPE? Explain the steps of your substitution.
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The expressions are
Substituting as indicated we get
We could factor P0 out of the last factor to get
`dPE = V0 * P0 * (T2 / T1 - 1) * ( T1 / T0 - 1)
which could be expressed more conceptually as
`dPE = P0 * V0 * ( (temp ratio state 2 to state 1) - 1) * ( (temp ratio state 1 to state 0) - 1)
The temperature ratios T2 / T1 and T1 / T0 are not difficult to understand. T1 is somewhere between T0 and T2. The higher T1, the greater the value of T1 / T0 and the less the value of T2 / T1. The closer T1 is to T0 the closer T1 / T0 is to 1, and the closer (T1 / T0 - 1) is to zero. The closer T1 is to T2 the closer T2 / T1 is to 1 and the closer (T2 / T1 - 1) is to zero.
So the two temperature ratios in the above expression, and they conflict in the sense that when one is greater, the other is less (the greater the state 1 temperature, the greater the state 1 to state 0 ratio, but the less the state 1 to state 2 ratio). If the temperature doesn't change between two states, then the temperature ratio is 1 and the corresponding factor (temp ratio - 1) will be zero. The result is no change in PE.
For example if T1 = T0, then T1 / T0 = 1 and (temp ratio state 1 to state 0 - 1) = (T1 / T0 - 1) = 1 - 1 = 0.
If T1 = T2, then T2 / T1 = 1 and (temp ratio state 2 to state 1 - 1) = (T2 / T1 - 1) = 1 - 1 = 0.
Thus our maximum PE change will occur at some intermediate temperature between T0 and T2.
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`q007. The expression you obtained at the end of the last question could be rearranged to read as follows:
`dPE = V0 * P0 * (T2 / T1 - 1) * (T1 / T0 - 1).
For T0 = 300 K, T1 = 360 K and T2 = 480 K, what are the values of (T2 / T1 - 1) and (T1 / T0 - 1)?
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T2 / T1 = 480 K / (360 K) = 1.33
T1 / T0 = 360 K / (300 K) = 1.2 so
and
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What do you get when you multiply these values of (T2 / T1 - 1) * ( T1 / T0 - 1)?
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The values are .33 and .2, and the product is therefore
(T2 / T1 - 1) * ( T1 / T0 - 1) = .33 * .2 = .066.
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For T0 and T2 as before, we investigate the effect of T1 on the product (T2 / T1 - 1) * (T1 / T0 - 1).
Suppose T1 = 320 K. Then what are the values of (T2 / T1 - 1) and (T1 / T0 - 1)? What therefore is the value of (T2 / T1 - 1) * (T1 / T0 - 1)?
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The calculation is similar to that done previously.
T1 / T0 = 320 K / (300 K) = 1.07, approx., and
T2 / T1 = 480 K / (320 K) = 1.5, so
(T2 / T1 - 1) * ( T1 / T0 - 1) = (1.5 - 1) * (1.07 - 1) = .5 * .07 = .035, very approximately.
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Answer the same questions for T1 = 305 K.
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T1 / T0 = 305 K / (300 K) = 1.017, approx., and
T2 / T1 = 480 K / (305 K) = 1.65, approx., so
(T2 / T1 - 1) * ( T1 / T0 - 1) = (1.65 - 1) * (1.017 - 1) = .65 * .017 = .012, very approximately.
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How do your results for T1 = 320 K compare to those for 305 K? What do you think happens to your quantities as T1 approaches 300 K?
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The values are .035 and .012.
The ratio T1 / T0 goes from 1.07 to 1.017, so that (T1 / T0 - 1) goes from .07 to .017. The value of (T2 / T1 - 1) increases from .5 to .65, but the percent decrease in the (T1 / T0 - 1) is much greater than the percent increase in (T2 / T1 - 1), so the product of these two factors decreases significantly (from .035 to .012 the decrease is nearly 70%).
As T1 continues to approach T0, the ratio T1 / T0 will approach 1 and (T1 / T0 - 1) will approach 1 - 1 = 0, so even though (T2 / T1 - 1) will continue to increase a bit, approaching a limit near 1.67, the product (T1 / T0 - 1) * (T2 / T1 - 1) will approach zero.
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Suppose T1 = 450 K. Then what are the values of (T2 / T1 - 1) and (T1 / T0 - 1)? What therefore is the value of (T2 / T1 - 1) * (T1 / T0 - 1)?
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Our calculations are again similar in nature to those done above. We will find that
(T2 / T1 - 1) * (T1 / T0 - 1) = (1.07 - 1) * (1.5 - 1) = .07 * .5 = .035.
This time the factor (T2 / T1 - 1) is the smaller.
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What do you think happens to your quantities as T1 approaches 480 K?
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T1 / T0 will approach 480 K / (300 K) = 1.60, so (T1 / T0 - 1) will approach .60.
However T2 / T1 will approach 480 K / (480 K) = 1, so (T2 / T1 - 1) will approach 1 - 1 = 0.
The product therefore approaches .60 * 0 = 0, and again we get no PE change.
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