How is pressure an intensive property?
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I've seen this question asked before but I can't find an answer to the specific point I'm troubled with. From the kinetic theory of gases, pressure results from molecules colliding with the walls of a container enclosing a gas, imparting a force upon the wall. Now, if we split the container into two halves, I am told that pressure remains the same on either side of the partition assuming the gas has uniform density throughout the container. But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
fluid-dynamics pressure
edited 1 hour ago
QuIcKmAtHs
2,4814828
asked 2 hours ago
user552217
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up vote
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I've seen this question asked before but I can't find an answer to the specific point I'm troubled with. From the kinetic theory of gases, pressure results from molecules colliding with the walls of a container enclosing a gas, imparting a force upon the wall. Now, if we split the container into two halves, I am told that pressure remains the same on either side of the partition assuming the gas has uniform density throughout the container. But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
fluid-dynamics pressure
edited 1 hour ago
QuIcKmAtHs
2,4814828
asked 2 hours ago
user552217
243
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've seen this question asked before but I can't find an answer to the specific point I'm troubled with. From the kinetic theory of gases, pressure results from molecules colliding with the walls of a container enclosing a gas, imparting a force upon the wall. Now, if we split the container into two halves, I am told that pressure remains the same on either side of the partition assuming the gas has uniform density throughout the container. But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
fluid-dynamics pressure
edited 1 hour ago
QuIcKmAtHs
2,4814828
asked 2 hours ago
user552217
243
I've seen this question asked before but I can't find an answer to the specific point I'm troubled with. From the kinetic theory of gases, pressure results from molecules colliding with the walls of a container enclosing a gas, imparting a force upon the wall. Now, if we split the container into two halves, I am told that pressure remains the same on either side of the partition assuming the gas has uniform density throughout the container. But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
fluid-dynamics pressure
fluid-dynamics pressure
edited 1 hour ago
QuIcKmAtHs
2,4814828
asked 2 hours ago
user552217
243
edited 1 hour ago
QuIcKmAtHs
2,4814828
asked 2 hours ago
user552217
243
edited 1 hour ago
QuIcKmAtHs
2,4814828
edited 1 hour ago
QuIcKmAtHs
2,4814828
edited 1 hour ago
QuIcKmAtHs
2,4814828
2,4814828
asked 2 hours ago
user552217
243
asked 2 hours ago
user552217
243
asked 2 hours ago
user552217
243
243
add a comment |
add a comment |
2 Answers
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If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
You are right that if we only halved the number of particles we would have a smaller pressure. But you have also halved the volume of the container. The fewer number of particles hits the walls more frequently due to the smaller volume. In other words, the number of particles goes down, but the number of collisions per particle goes up. The two effects cancel out, leading to the same pressure as before you put in the partition.
answered 2 hours ago
Aaron Stevens
8,46931239
add a comment |
up vote
2
down vote
But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
The pressure is not dependent on the number of molecules alone. You can simply examine the ideal gas law: $PV=nRT$. If the temperature is constant then reducing both $n$ and $V$ by half leaves pressure unchanged.
answered 2 hours ago
Dale
4,4891623
1
Physical answer vs. mathematical answer. Let's see who wins :p
– Aaron Stevens
2 hours ago
Usually physical does, but the mathematical one was so simple in this case.
– Dale
2 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
You are right that if we only halved the number of particles we would have a smaller pressure. But you have also halved the volume of the container. The fewer number of particles hits the walls more frequently due to the smaller volume. In other words, the number of particles goes down, but the number of collisions per particle goes up. The two effects cancel out, leading to the same pressure as before you put in the partition.
answered 2 hours ago
Aaron Stevens
8,46931239
add a comment |
up vote
3
down vote
If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
You are right that if we only halved the number of particles we would have a smaller pressure. But you have also halved the volume of the container. The fewer number of particles hits the walls more frequently due to the smaller volume. In other words, the number of particles goes down, but the number of collisions per particle goes up. The two effects cancel out, leading to the same pressure as before you put in the partition.
answered 2 hours ago
Aaron Stevens
8,46931239
add a comment |
up vote
3
down vote
up vote
3
down vote
If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
You are right that if we only halved the number of particles we would have a smaller pressure. But you have also halved the volume of the container. The fewer number of particles hits the walls more frequently due to the smaller volume. In other words, the number of particles goes down, but the number of collisions per particle goes up. The two effects cancel out, leading to the same pressure as before you put in the partition.
answered 2 hours ago
Aaron Stevens
8,46931239
If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
You are right that if we only halved the number of particles we would have a smaller pressure. But you have also halved the volume of the container. The fewer number of particles hits the walls more frequently due to the smaller volume. In other words, the number of particles goes down, but the number of collisions per particle goes up. The two effects cancel out, leading to the same pressure as before you put in the partition.
answered 2 hours ago
Aaron Stevens
8,46931239
answered 2 hours ago
Aaron Stevens
8,46931239
answered 2 hours ago
Aaron Stevens
8,46931239
answered 2 hours ago
Aaron Stevens
8,46931239
8,46931239
add a comment |
add a comment |
up vote
2
down vote
But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
The pressure is not dependent on the number of molecules alone. You can simply examine the ideal gas law: $PV=nRT$. If the temperature is constant then reducing both $n$ and $V$ by half leaves pressure unchanged.
answered 2 hours ago
Dale
4,4891623
1
Physical answer vs. mathematical answer. Let's see who wins :p
– Aaron Stevens
2 hours ago
Usually physical does, but the mathematical one was so simple in this case.
– Dale
2 hours ago
add a comment |
up vote
2
down vote
But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
The pressure is not dependent on the number of molecules alone. You can simply examine the ideal gas law: $PV=nRT$. If the temperature is constant then reducing both $n$ and $V$ by half leaves pressure unchanged.
answered 2 hours ago
Dale
4,4891623
1
Physical answer vs. mathematical answer. Let's see who wins :p
– Aaron Stevens
2 hours ago
Usually physical does, but the mathematical one was so simple in this case.
– Dale
2 hours ago
add a comment |
up vote
2
down vote
up vote
2
down vote
But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
The pressure is not dependent on the number of molecules alone. You can simply examine the ideal gas law: $PV=nRT$. If the temperature is constant then reducing both $n$ and $V$ by half leaves pressure unchanged.
answered 2 hours ago
Dale
4,4891623
But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
The pressure is not dependent on the number of molecules alone. You can simply examine the ideal gas law: $PV=nRT$. If the temperature is constant then reducing both $n$ and $V$ by half leaves pressure unchanged.
answered 2 hours ago
Dale
4,4891623
answered 2 hours ago
Dale
4,4891623
answered 2 hours ago
Dale
4,4891623
answered 2 hours ago
Dale
4,4891623
4,4891623
1
Physical answer vs. mathematical answer. Let's see who wins :p
– Aaron Stevens
2 hours ago
Usually physical does, but the mathematical one was so simple in this case.
– Dale
2 hours ago
add a comment |
1
Physical answer vs. mathematical answer. Let's see who wins :p
– Aaron Stevens
2 hours ago
Usually physical does, but the mathematical one was so simple in this case.
– Dale
2 hours ago
1
1
Physical answer vs. mathematical answer. Let's see who wins :p
– Aaron Stevens
2 hours ago
Physical answer vs. mathematical answer. Let's see who wins :p
– Aaron Stevens
2 hours ago
Usually physical does, but the mathematical one was so simple in this case.
– Dale
2 hours ago
Usually physical does, but the mathematical one was so simple in this case.
– Dale
2 hours ago
add a comment |
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