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Say a gas sample is occupying 300 cm3 is moved to a container that with a volume of 200 cm3. What changes would happen to the following values?
a. average kinetic energy of the gas molecules
b. the pressure of the gas
c. the average speed of the gas molecules
d. the number of collisions of a gas molecule with another gas molecule

2007-03-13 11:42:17 · 3 answers · asked by San Fran Kid 2 in Science & Mathematics Chemistry

3 answers

Assuming that it's an adiabatic process, meaning that work has to be done to compress it (you know, it could have been cooled, in which cases the answers would be different), then

a) Goes UP, because there's now more internal energy in the gas because of the work done to it
b) Goes UP, because both work has been done to it, and p = T/V, and T went up while V went down
c) Goes UP because of a)
d) Goes UP because of c), while V went down

On the other hand, if suddenly 300 cm3 of gas found itself confned into a 200 cm3 by some miracle, all other things being the same, then:

a) Remains the SAME
b) Goes UP, because p = T/V, and V went down
c) Remains the SAME because of a)
d) Goes UP, because of a), while V went down.

And finally, what if the gas was COOLED so that the volume was shrunken? (Imagine that it was a 300 cm3 balloon placed in a nitrogen bath).

a) Goes DOWN, because temperature has gone down
b) Remains the SAME, because the size of the volume was set by the pressure of environmental pressure, i.e., the nitrogen bath
c) Goes DOWN, because of a)
d) This one needs to be worked out, so I'll leave this to you as an exercise.

Confused? Hey, it's thermodynamics!

2007-03-13 12:01:46 · answer #1 · answered by Scythian1950 7 · 0 0

when the volume decreases pressure will increase
kE will increase as well so the speed will increase since ke=1/2mv^2
# collisions will increase

2007-03-13 11:50:15 · answer #2 · answered by essebful 2 · 0 0

i think the answer is a. because then the molecules have less room to move so they would have less kenetic energy.

2007-03-13 11:51:56 · answer #3 · answered by Anonymous · 0 0

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