Please solve parts B and C Par
Please solve parts B and C
Part A
What is the average time required for H2 to travel 1.00 m at 298K and 1 atm?
Express your answer with the appropriate units.
| t = |
5.66×10−4 s |
Correct
Part B
How much longer does it take N2 to travel 1.00 m, on average,relative to H2 under these same conditions?
Express your answer with the appropriate units.
| t = |
3.73s |
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Part C
(Challenging) What fraction of N2 particles will require morethan this average time to travel 1.00 m? Answering this questionwill require evaluating a definite integral of the speeddistribution, which requires using numerical methods such asSimpson’s rule.
|
|
|||
| f = |
3.6 |
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Answer:
b)
Compare Velocities
i)
Apply the RMS equation
RMS =sqrt(3*RT/MW)
where, R = ideal gas constant, T = temperature, MW moalr mas inkg/mol
MW of H2 = 2g/mol –> change to kg –> 2*10^-3 kg/mol
RMS= sqrt(3*8.314*298/ (2*10^-3))
RMS = 1927.78 m/s
ii)
Apply the RMS equation
RMS =sqrt(3*RT/MW)
where, R = ideal gas constant, T = temperature, MW moalr mas inkg/mol
MW of N2 = 28 g/mol –> change to kg –> 28*10^-3kg/mol
RMS= sqrt(3*8.314*298/ (28*10^-3))
RMS = 515.2 m/s
Copmare V:
VN2 / VH2 = 1927.78 /515.2 = 3.741 times longer
if unit of S is required
time H2 = D/V = (1)/(1927.78 ) = 0.0005187 s
time (N2) = 3.741 *0.0005187 s = 0.0019404
dT = 0.0019404-0.0005187 = 0.0014217 s
cfraction of N2
According to maxwell distribution –> fraction
apply
P(V< Vavg) = integral (0 to Vavg) P(v)dV
P(V< Vavg) = 4PI*(m/(2PI*k*t)^1.5 * integral (0 to avg) v^2 *exp(-mv^2)/(2Kt) dV
now
let x^2 = mv^2 /(2kT)
x = v*sqrt(m/(2kT))
solving
V = sqrt(2kT/m) * x
get differentials
dv = Vavg * sqrt(m/(2kT))
developing this:
Px = 4/(sqrtPI) * (PI*erf(2/sqrt(PI)) – 4exp(-4/PI)) /(4*sqrt(PI)
note that (PI*erf(2/sqrt(PI)) – 4exp(-4/PI)) / (4*sqrt(PI) =0.2405
then
PX = 4/sqrt(3.1416) * 0.2405
PX = 0.5427
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