Question

Please solve parts B and C

Part A

What is the average time required for H2 to travel 1.00 m at 298 K 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 more than this average time to travel 1.00 m? Answering this question will require evaluating a definite integral of the speed distribution, which requires using numerical methods such as Simpson's rule.

f =	3.6

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Answer 1

b)

Compare Velocities

i)

Apply the RMS equation

RMS =sqrt(3*RT/MW)

where, R = ideal gas constant, T = temperature, MW moalr mas in kg/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 in kg/mol

MW of N2 = 28 g/mol --> change to kg --> 28*10^-3 kg/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