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anonymous
 one year ago
According to the Kinetic Molecular Theory, as the volume of an enclosed gas increases:
the number of particle impacts per unit area increases
the temperature must change
the total number of particle impacts increase
the number of particle impacts per unit area decreases
anonymous
 one year ago
According to the Kinetic Molecular Theory, as the volume of an enclosed gas increases: the number of particle impacts per unit area increases the temperature must change the total number of particle impacts increase the number of particle impacts per unit area decreases

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Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[pV = nRT\] let's keep pressure constant, and number of moles constant. nR = constant \[V_{1} = T_{1} nR \] \[\frac{ V_{1} }{ T_{1} } = nR = \frac{ V_{2} }{ T_{2} }\] \[\frac{ V_{1} }{ T_{1} } = \frac{ V_{2} }{ T_{2} }\] If we plot volume and temperature we get a straight line. dw:1444417451411:dw if the volume goes up the temperature goes up too.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1To test the last claim we need to keep the temperature constant to study pressure and volume. \[pV = nRT \] \[P_{1}V_{1} = nR = k = P_{2}V_{2}\] IF increase the volume or make V_2 bigger, for the two to equal echoer, the pressure must go down. pressure translates to how many times the particles hit the walls of the container fewer times. so that's also true.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you sir much obliged

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Select the TWO variables that are held constant when testing Boyle's Law in a manometer. pressure temperature volume of gas mass of gas atmospheric pressure?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i cant decide which one to pick

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@23188 no problem, I provided you an explanation for your other question too.
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