Can the van der Waals coefficients be negative in the van der Waals equation for real gases? ...
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Can the van der Waals coefficients be negative in the van der Waals equation for real gases?
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2019 Moderator Election Q&A - Question CollectionHigh pressure modification to the van der Waals equationDoes the van der Waals equation remain valid when repulsive intermolecular forces dominate?Van der Waals real gas equationvan der Waals equation of state constantsDeriving an alternative expression for the Van der Waals equation using given parametersDerivation of the van der Waals equationValidity of van der Waals equationCalculating Compressibility factor from the Van der Waals' Gas equationvan der Waals equation for deviation of gases from ideal behaviorVan der Waals forces
$begingroup$
A gas described by van der Waals equation has the pressure that is lower than the pressure exerted by the same gas behaving ideally. True or false?
My approach
$$P(text{ideal}) = P(text{real}) + P(text{changes due to intermolecular forces})$$
So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.
Doubts
The answer to the question is given as True. So I want to understand what is wrong about my approach.
Since the intermolecular forces depend on $a$, for attractive nature $a$ is positive. Is $a$ negative for repulsive nature of forces?
physical-chemistry gas-laws
edited 6 hours ago
andselisk
19.6k666128
asked 7 hours ago
GroverkssGroverkss
777
$endgroup$
add a comment |
$begingroup$
A gas described by van der Waals equation has the pressure that is lower than the pressure exerted by the same gas behaving ideally. True or false?
My approach
$$P(text{ideal}) = P(text{real}) + P(text{changes due to intermolecular forces})$$
So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.
Doubts
The answer to the question is given as True. So I want to understand what is wrong about my approach.
Since the intermolecular forces depend on $a$, for attractive nature $a$ is positive. Is $a$ negative for repulsive nature of forces?
physical-chemistry gas-laws
edited 6 hours ago
andselisk
19.6k666128
asked 7 hours ago
GroverkssGroverkss
777
$endgroup$
$begingroup$
Upon further thought, I have an explanation of why what I am thinking is wrong. The constant 'a' only describes attractive forces while the constant 'b' describes repulsive forces. So by definition 'a' cannot be negative. Can anyone tell me if this is correct? I'm still unsure about the 1st doubt.
$endgroup$
– Groverkss
7 hours ago
$begingroup$
Rather, b does not address repulsive forces, but the volume of molecules that is neglected in the ideal gas equation. As consequence, the gas has higher pressure, as collisions with volume walls are more frequent.
$endgroup$
– Poutnik
6 hours ago
$begingroup$
@Poutnik Isn't repulsive forces and volume neglected the same thing? I think that volume occupied is due to repulsive forces between atoms. I may be wrong, can you clarify?
$endgroup$
– Groverkss
6 hours ago
2
$begingroup$
No, it is not due repulsive forces. Neither for ideal gas, neither for real gas. Bigger own volume = lower fre volume = more frequent wall collisions.
$endgroup$
– Poutnik
6 hours ago
add a comment |
$begingroup$
A gas described by van der Waals equation has the pressure that is lower than the pressure exerted by the same gas behaving ideally. True or false?
My approach
$$P(text{ideal}) = P(text{real}) + P(text{changes due to intermolecular forces})$$
So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.
Doubts
The answer to the question is given as True. So I want to understand what is wrong about my approach.
Since the intermolecular forces depend on $a$, for attractive nature $a$ is positive. Is $a$ negative for repulsive nature of forces?
physical-chemistry gas-laws
edited 6 hours ago
andselisk
19.6k666128
asked 7 hours ago
GroverkssGroverkss
777
$endgroup$
A gas described by van der Waals equation has the pressure that is lower than the pressure exerted by the same gas behaving ideally. True or false?
My approach
$$P(text{ideal}) = P(text{real}) + P(text{changes due to intermolecular forces})$$
So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.
Doubts
The answer to the question is given as True. So I want to understand what is wrong about my approach.
Since the intermolecular forces depend on $a$, for attractive nature $a$ is positive. Is $a$ negative for repulsive nature of forces?
physical-chemistry gas-laws
physical-chemistry gas-laws
edited 6 hours ago
andselisk
19.6k666128
asked 7 hours ago
GroverkssGroverkss
777
edited 6 hours ago
andselisk
19.6k666128
asked 7 hours ago
GroverkssGroverkss
777
edited 6 hours ago
andselisk
19.6k666128
edited 6 hours ago
andselisk
19.6k666128
edited 6 hours ago
andselisk
19.6k666128
19.6k666128
asked 7 hours ago
GroverkssGroverkss
777
asked 7 hours ago
GroverkssGroverkss
777
asked 7 hours ago
GroverkssGroverkss
777
777
$begingroup$
Upon further thought, I have an explanation of why what I am thinking is wrong. The constant 'a' only describes attractive forces while the constant 'b' describes repulsive forces. So by definition 'a' cannot be negative. Can anyone tell me if this is correct? I'm still unsure about the 1st doubt.
$endgroup$
– Groverkss
7 hours ago
$begingroup$
Rather, b does not address repulsive forces, but the volume of molecules that is neglected in the ideal gas equation. As consequence, the gas has higher pressure, as collisions with volume walls are more frequent.
$endgroup$
– Poutnik
6 hours ago
$begingroup$
@Poutnik Isn't repulsive forces and volume neglected the same thing? I think that volume occupied is due to repulsive forces between atoms. I may be wrong, can you clarify?
$endgroup$
– Groverkss
6 hours ago
2
$begingroup$
No, it is not due repulsive forces. Neither for ideal gas, neither for real gas. Bigger own volume = lower fre volume = more frequent wall collisions.
$endgroup$
– Poutnik
6 hours ago
add a comment |
$begingroup$
Upon further thought, I have an explanation of why what I am thinking is wrong. The constant 'a' only describes attractive forces while the constant 'b' describes repulsive forces. So by definition 'a' cannot be negative. Can anyone tell me if this is correct? I'm still unsure about the 1st doubt.
$endgroup$
– Groverkss
7 hours ago
$begingroup$
Rather, b does not address repulsive forces, but the volume of molecules that is neglected in the ideal gas equation. As consequence, the gas has higher pressure, as collisions with volume walls are more frequent.
$endgroup$
– Poutnik
6 hours ago
$begingroup$
@Poutnik Isn't repulsive forces and volume neglected the same thing? I think that volume occupied is due to repulsive forces between atoms. I may be wrong, can you clarify?
$endgroup$
– Groverkss
6 hours ago
2
$begingroup$
No, it is not due repulsive forces. Neither for ideal gas, neither for real gas. Bigger own volume = lower fre volume = more frequent wall collisions.
$endgroup$
– Poutnik
6 hours ago
$begingroup$
Upon further thought, I have an explanation of why what I am thinking is wrong. The constant 'a' only describes attractive forces while the constant 'b' describes repulsive forces. So by definition 'a' cannot be negative. Can anyone tell me if this is correct? I'm still unsure about the 1st doubt.
$endgroup$
– Groverkss
7 hours ago
$begingroup$
Upon further thought, I have an explanation of why what I am thinking is wrong. The constant 'a' only describes attractive forces while the constant 'b' describes repulsive forces. So by definition 'a' cannot be negative. Can anyone tell me if this is correct? I'm still unsure about the 1st doubt.
$endgroup$
– Groverkss
7 hours ago
$begingroup$
Rather, b does not address repulsive forces, but the volume of molecules that is neglected in the ideal gas equation. As consequence, the gas has higher pressure, as collisions with volume walls are more frequent.
$endgroup$
– Poutnik
6 hours ago
$begingroup$
Rather, b does not address repulsive forces, but the volume of molecules that is neglected in the ideal gas equation. As consequence, the gas has higher pressure, as collisions with volume walls are more frequent.
$endgroup$
– Poutnik
6 hours ago
$begingroup$
@Poutnik Isn't repulsive forces and volume neglected the same thing? I think that volume occupied is due to repulsive forces between atoms. I may be wrong, can you clarify?
$endgroup$
– Groverkss
6 hours ago
$begingroup$
@Poutnik Isn't repulsive forces and volume neglected the same thing? I think that volume occupied is due to repulsive forces between atoms. I may be wrong, can you clarify?
$endgroup$
– Groverkss
6 hours ago
2
2
$begingroup$
No, it is not due repulsive forces. Neither for ideal gas, neither for real gas. Bigger own volume = lower fre volume = more frequent wall collisions.
$endgroup$
– Poutnik
6 hours ago
$begingroup$
No, it is not due repulsive forces. Neither for ideal gas, neither for real gas. Bigger own volume = lower fre volume = more frequent wall collisions.
$endgroup$
– Poutnik
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This question requires a simplistic notion of real gas behavior.
The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction $b$ is negative in the equation and the pressure correction, $a$ is positive. The formula is
$$(P + a/V_mathrm{m}^2)(V_mathrm{m} -b) = RT$$
If you look at a table of van der Waals constants all the a and b terms are positive. Thus the volume calculated using the van der Waals equation will always be less than the volume calculated using the ideal gas equation.
The rest of the story...
The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)
answered 6 hours ago
MaxWMaxW
16k22261
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This question requires a simplistic notion of real gas behavior.
The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction $b$ is negative in the equation and the pressure correction, $a$ is positive. The formula is
$$(P + a/V_mathrm{m}^2)(V_mathrm{m} -b) = RT$$
If you look at a table of van der Waals constants all the a and b terms are positive. Thus the volume calculated using the van der Waals equation will always be less than the volume calculated using the ideal gas equation.
The rest of the story...
The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)
answered 6 hours ago
MaxWMaxW
16k22261
$endgroup$
add a comment |
$begingroup$
This question requires a simplistic notion of real gas behavior.
The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction $b$ is negative in the equation and the pressure correction, $a$ is positive. The formula is
$$(P + a/V_mathrm{m}^2)(V_mathrm{m} -b) = RT$$
If you look at a table of van der Waals constants all the a and b terms are positive. Thus the volume calculated using the van der Waals equation will always be less than the volume calculated using the ideal gas equation.
The rest of the story...
The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)
answered 6 hours ago
MaxWMaxW
16k22261
$endgroup$
add a comment |
$begingroup$
This question requires a simplistic notion of real gas behavior.
The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction $b$ is negative in the equation and the pressure correction, $a$ is positive. The formula is
$$(P + a/V_mathrm{m}^2)(V_mathrm{m} -b) = RT$$
If you look at a table of van der Waals constants all the a and b terms are positive. Thus the volume calculated using the van der Waals equation will always be less than the volume calculated using the ideal gas equation.
The rest of the story...
The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)
answered 6 hours ago
MaxWMaxW
16k22261
$endgroup$
This question requires a simplistic notion of real gas behavior.
The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction $b$ is negative in the equation and the pressure correction, $a$ is positive. The formula is
$$(P + a/V_mathrm{m}^2)(V_mathrm{m} -b) = RT$$
If you look at a table of van der Waals constants all the a and b terms are positive. Thus the volume calculated using the van der Waals equation will always be less than the volume calculated using the ideal gas equation.
The rest of the story...
The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)
answered 6 hours ago
MaxWMaxW
16k22261
edited 4 hours ago
answered 6 hours ago
MaxWMaxW
16k22261
answered 6 hours ago
MaxWMaxW
16k22261
answered 6 hours ago
MaxWMaxW
16k22261
16k22261
add a comment |
add a comment |
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$begingroup$
Upon further thought, I have an explanation of why what I am thinking is wrong. The constant 'a' only describes attractive forces while the constant 'b' describes repulsive forces. So by definition 'a' cannot be negative. Can anyone tell me if this is correct? I'm still unsure about the 1st doubt.
$endgroup$
– Groverkss
7 hours ago
$begingroup$
Rather, b does not address repulsive forces, but the volume of molecules that is neglected in the ideal gas equation. As consequence, the gas has higher pressure, as collisions with volume walls are more frequent.
$endgroup$
– Poutnik
6 hours ago
$begingroup$
@Poutnik Isn't repulsive forces and volume neglected the same thing? I think that volume occupied is due to repulsive forces between atoms. I may be wrong, can you clarify?
$endgroup$
– Groverkss
6 hours ago
2
$begingroup$
No, it is not due repulsive forces. Neither for ideal gas, neither for real gas. Bigger own volume = lower fre volume = more frequent wall collisions.
$endgroup$
– Poutnik
6 hours ago