Question

At what temperature would the most probale speed of CO2 molecules be twice that at 323K?

Answer 1

MPV = sqrt 2RT/M

Answer 2

@UnkleRhaukus plz help to figure out this

Answer 3

@eigenschmeigen

Answer 4

@ujjwal @UnkleRhaukus any idea

Answer 5

\[v_{\text{rms}}=\sqrt{\frac{3RT}{M}}\]

Answer 6

most probale speed

Answer 7

plzz read the quest once more

Answer 8

\[M_{\text{CO}_2}=44[\text{amu}]\]

Answer 9

http://www.physicsforums.com/showthread.php?t=274201

Answer 10

you just need to use the relation \[v=k\sqrt T\] at two different absolute temperatures.
Here k is constant.
\[v_1 =k\sqrt {323}\]\[v_2=k\sqrt {T_2}\]But \(v_2=2v_1\)
so, we have \[\frac{v_1}{2v_1}=\sqrt{\frac{323}{T_2}}\]so, \(T_2\)= 1292 K=1019 degrees.

Answer 11

yup u r correct

Answer 12

Let me try to undersstand

Answer 13

i got it thxxxx @ujjwal

Answer 14

wc. :)

Answer 15

Plzz look out my next question! @ujjwal