Question

A gas occupies 1560 cm3 at 285 K. To what temp must the gas be lowered to, if it is to occupy 25.0 cm3? Assume a constant pressure.

Answer 1

would it be set up like this 285 K/1560cm3 X ?/25.0cm3

Answer 2

yeah, you can set up it either way

\[\frac{ T _{1} }{ V _{1} }=\frac{ T _{2} }{ V _{2} } or \frac{ V _{1} }{ T _{1} }=\frac{ V _{2} }{ T _{2} } \]

Answer 3

so then it would be 444600/1560=285K is that the final answer???

Answer 4

I messed up somewhere

Answer 5

285/1560=x/25 -> x=4.5673076923076923 K

Answer 6

0.1826K??

Answer 7

thats not what i got

Answer 8

so you don't take 4.57307/25.0???

Answer 9

no, you're solving for the T when the gas occupies 25 cm^3, which x

Answer 10

which is*

Answer 11

multiply??? 4.567307 by 25.0cm3

Answer 12

would i take 1560/25.0=6.24

Answer 13

nonono,
This is the set up:

285 K/1560cm3 =T/25.0cm3 -> T=285 K*25.0cm3/1560cm3= 4.57 K

Answer 14

thank you again ;D

Answer 15

hah no problem !

Answer 16

I wish it was that easy for me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ;D

Answer 17

tHANKYOU

Answer 18

haha practice is all it takes :P

you're on your own now :S, though, i gotta go study got a test tmorw

Answer 19

Good luck not that you'll need it

Answer 20

your a life saver