IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

joyfrancis

calculate the root mean square velocity of ozone kept in a closed vessel at 20^oC and 82 cm Hg pressure.

(here the pressure is not small so will the behavior of the gas deviate from ideal gas behavior, how do we approach such problems?)

TigerEyes

i think we just need to apply root(3RT/M) also gases behave ideally at high temp & low pressure .. the given pressure can be assumed to be low compared to 760 mm Hg pressure(atm pressure) ........

if the the gas was to be considered non ideal then the question will surely give the compressibilty factor then apply this eqn to calculate rms velocity > root(3ZPV/M) , where Z is the compressibilty factor at that temp.

thnx

joyfrancis

Give me the answer please!

TigerEyes

ans is : 12.34 m/s.

joyfrancis

I got the same answer but it's incorrect. The correct answer is 3.9*10^2 cm/sec

TigerEyes

hey sorry for that incorrect answer, the approach ofthe problem depends on the choices given and the data given , here the data is temp & pressure, we dont have a direct formula relating rms velocity to pressure & temp , so we need to use our old root (3PV/M)

now here temp = 293K (20+273) & pressure = 82m Hg.

we know P1V1/T1 = P2V2/T2

find volume V2 when P1 = 76 cm of Hg, V1= 22400 cm^3 & T1 = 273 K & P2=82cm of Hg, T2 = 293 K

V2 comes out to be = 201.3 * 10^3 cm^3

now

v= root(3P2V2/48)

=396CM/SEC

arun-rashi

according to given data V rms =root 3pv/m here first calculate volume at given pressure and temperature by P1V1/T1 =P2V2/T2 After this you get the answer.

payoj

sqr root (3RT/M) will give the same answer as above, i.e.390 m/s.

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