Resource Lesson
State Variables
Printer Friendly Version
When examining PV graphs, state variables are ones that always maintain a constant value at a given position on the graph. These variables are:
P, V, T, U, and S, where
P is pressure
V is volume
T is absolute temperature (K)
U is the gas' internal energy
S is entropy
The change in the value of a state variable (
Δ
P,
Δ
V,
Δ
S,
Δ
U, and
Δ
T) is path-independent which means that we only need to be concerned with the value at the final position and the value at the initial position -just like we did with
displacement vectors
.
Temperature
If n, the amount of gas, does not change, then according to the ideal gas law, PV ∝ T.
We can use this relationship to determine how the temperatures at these different positions on a compare by comparing their "PV" products. For example, the "PV" product of point B equals 5 while the "PV" product of C only equals 1. This would tell us that the absolute temperature of position B is 5-times as great as that at C. While the "PV" products at A and I are both 2.4 telling us that these two positions have the same absolute temperature.
Internal Energy
The next state variable in our list is U, the internal energy of the gas. For a gas, the change in its internal energy, ΔU, is directly proportional to the change in its temperature measured in Kelvin.
ΔU = (3/2)nRΔT for a monatomic gas
ΔU = (5/2)nRΔT for a diatomic gas
This means that ΔU only reflects a change in the kinetic energy of the gas molecules. Remember that the potential energy can not change except when there is a phase change: liquid to solid or liquid to vapor, and we are only working with gases.
Using the
kinetic theory of gases
, we can also state that the internal energy at a given position on a PV graph equals
U = N(
),
where the average translational kinetic energy is given by the relationship
.
Using this information, once again assuming that the amount and type of gas is constant at each point on our PV graph, then the internal energy of each position on the graph follows in the same order as their absolute temperatures.
Entropy
The final state variable is entropy, S. This variable applies to the amount of disorder in the system - that is, the probability of being able to pinpoint the location of a particular molecule at a given moment. Entropy increase with the addition of heat to a system. So entropy would increase when an ice cube at 0ºC melts into a puddle of water at 0ºC, or a cup of 100ºC water boils to 100ºC steam. Logically, the entropy of gases is greater than that of liquids which is greater than that of solids. The formula for entropy at a given temperature is S = Q/T. On our PV graph, the change in entropy would be indicated by a transition that involves either the addition or removal of heat.
Calculating the thermal equilibrium temperature for two gas samples
The Zeroth Law of Thermodynamics states that if System A and System B are in thermal equilibrium and System B and System C are in thermal equilibrium, then System A and System C are also in thermal equilibrium. This is very similar to the transitive property of algebra: If A = B and B = C, then A = C.
If two gas samples are separated by a partition, then the Second Law of Thermodynamics tells us that heat will ONLY flow naturally from the high temperature gas to the low temperature gas. And, that the heat will cease to flow when the two samples are in thermal equilibrium. If this situation, system B would be the partition between the two gas samples.
Let's do an example of calculating thermal equilibrium. Let sample A be 0.1 moles of helium gas at 200K and sample C be 0.2 moles of argon at 400K. Since we know that heat is the flow of internal energy from a high temperature "object" to a lower temperature "object" we should first determine how much total internal energy is initially present in our system.
Both helium and argon are noble gases, or monatomic gases. Therefore we can calculate their initial internal energies using the formula U = 3/2nRT.
U
_{He}
= (3/2)(0.1)(8.314)(200) = 249.4 J
U
_{Ar}
= (3/2)(0.2)(8.314)(400) = 997.7 J
Therefore, the total internal energy present in our system is 249.4 + 997.7 = 1247.1 J
Since energy is conserved, our process will not alter the total amount of internal energy, but will redistribute the energy as the two gas samples reach thermal equilibrium. Remember, that the transfer of heat is through the membrane (partition) so the molecules on both sides (A|B and B|C) will be undergoing collisions that transfer energy,
Upon reaching thermal equilibrium, all gas molecules will reach the same temperature and therefore have the identical amount of KE.
N
_{He}
KE
_{He}
= N
_{Ar}
KE
_{Ar}
= (N
_{He}
+ N
_{Ar}
)KE
_{final}
The amount of heat that flowed from the hotter argon gas to the cooler helium gas was
(831.4 - 997.7) = -166.3 J
and the amount of heat that flowed into the cooler helium gas was
(415.7 - 249.4) = +166.3 J.
To find the common final temperature, solve the formula for internal energy, U, for T and substitute in the final values for either helium or argon. The calculation for helium is given below. Test yourself and determine the final temperature for argon and see that the two temperatures are the same.
If the one or the other of the gases had been diatomic, U = (5/2)nRT, or polyatomic, U = (7/2)nRT.
Refer to the following information for the next six questions.
In each of the following examples, you may use the ideal gas law to determine the relationships and/or the equation for internal energy.
Two cylinders, that are NOT in thermal contact, each have the same number of molecules. Cylinder A has a volume of 2 m
^{3}
and a temperature of 300K while cylinder B has a volume of 3.0 m
^{3}
and a temperature of 450K. Which container has the greater gauge pressure?
A
B
neither, they have the same gauge pressure
The pressure and volume values of a container of gas are tracked as the temperature of the gas is changed. Which combination presents the greater temperature of the confined gas?
P = 3 atm, V = 1 L
P = 2.5 atm, V = 1.2 L
P = 1.5 atm, V = 2.5 L
all three represent the same temperature
The following three cylinders contain monatomic, ideal gases. You are not given the volume of any of the containers which could be or could not be the same. Determine which container's gas has the greatest internal energy,
n = 0.1 moles, T = 200K, P = 1 atm
n = 0.2 moles, T = 150K, P = 2.5 atm
n = 0.3 moles, T = 300K, P = 4 atm
all three containers hold gases with equal amounts of internal energy
Which of the following containers holds an ideal has at the highest temperature?
P = 3 atm, V = 1 m
^{3}
, n = 0.1 moles
P = 1.5 atm, V = 2 m
^{3}
, n = 0.2 moles
P = 1.25 atm, V = 2.4 m
^{3}
, n = 0.3 moles
they all three are at the same temperature
Which of the following containers holds an ideal gas at the highest temperature?
V = 1 m
^{3}
, n = 0.4 moles, U = 60 J
V = 2 m
^{3}
, n = 0.2 moles, U = 30 J
V = 1.5 m
^{3}
, n = 0.3 moles, U = 45 J
all containers are at the same temperature
Which of the following containers holds an ideal gas under the greatest pressure?
V = 1.5 m
^{3}
, n = 0.3 moles, U = 45 J
V = 1 m
^{3}
, n = 0.4 moles, U = 60 J
V = 2 m
^{3}
, n = 0.2 moles, U = 30 J
all containers are at the same pressure
Special thanks to Randall Knight for his approach to calculating the final internal energies of two containers in thermal equilibrium and to Thomas O'Kuma, David Maloney, and Curtis Hieggelke for their insights into the relationships between pressure, volume, number of gas molecules and internal energy in their ranking tasks.
Related Documents
Lab:
Labs -
A Sample Heat Engine
Resource Lesson:
RL -
2nd Law of Thermodynamics and Entropy
RL -
Heat Cycles
RL -
Thermodynamic Processes
Worksheet:
CP -
Thermodynamics
WS -
Heat Cycles
PhysicsLAB
Copyright © 1997-2023
Catharine H. Colwell
All rights reserved.
Application Programmer
Mark Acton