0.04 moles of an ideal gas (molecular diameter = 4.0 x 10^-10 m and mass = 32 g/

Question

0.04 moles of an ideal gas (molecular diameter = 4.0 x 10^-10 m and mass = 32 g/mole) occupies

a 1 liter (0.001 m^3) container at 1 atm of pressure and at room temperature (298 K).

(A). Calculate the mean speed (c) of the gas in the container.
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(B). Calculate the number of collisions a single molecule makes in one second.

(C). Calculate the amount of time (in seconds) that occurs in between each collision.

(D). Calculate the average mean free path of the gas.

(E). Calculate the binary collision frequency in units of Molar per second (M/s).

(F). If every collision between two ideal gas molecules causes formation of one product molecule, what is an upper limit for the rate of this reaction at 298 K (in units of Molar/s), explain?

Explanation / Answer

a)mean speed = sqrt(8RT/pi*M) = sqrt(8*8.314*298/3.14*0.032) = 444.04 m/s

b)number of collisions a single molecule makes in one second. =

vel/mean free path=444.04/(5.8428*10^-8)=75.99*10^8 hz

c) amount of time (in seconds) that occurs in between each collision.=

mean free path/vel=1.315 *10^-10 s

d) Nv =no.of molecules per unit volume = 0.04*6.023*10^23/0.001 = 2.4092*10^25

average mean free path of the gas. = 1/(1.414*pi*d^2*Nv) = 1/(1.414*pi*16*10^-20*2.4092*10^25) = 5.8428*10^-8 m = 584.28 A

e)

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