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?






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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







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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







      share|cite|improve this question















      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






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      QuIcKmAtHs

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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.






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            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.






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              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











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            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.






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              up vote
              3
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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.






              share|cite|improve this answer























                up vote
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                up vote
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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.






                share|cite|improve this answer













                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Aaron Stevens

                8,46931239




                8,46931239






















                    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.






                    share|cite|improve this answer

















                    • 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















                    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.






                    share|cite|improve this answer

















                    • 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













                    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.






                    share|cite|improve this answer













                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    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














                    • 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


















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