14.11 For 1.00 mol of O2 at 300 K and 1.00 atm, calculate (a) the number of mole

Question

14.11 For 1.00 mol of O2 at 300 K and 1.00 atm, calculate (a) the number of molecules whose speed lies in the range 500.000 to 500.001 m/s (because this speed interval is very small, the distribution function changes only very slightly over this interval, and the interval can be considered infinitesimal); (b) the number of molecules with v, in the range 150.000 to 150.001 m/s; (c) the number of molecules that simultaneously have v, in the range 150.000 to 150.001 m/s and have v, in the range 150.000 to 150.001 m/s. 14.12 For CH4(g) at 300 K and 1 bar, calculate the probabil- ity that a molecule picked at random has its speed in the range

Explanation / Answer

14.11 For O2, (a) number of molecules = (4piv^2)(M/(2piRT)^3/2.(e^(-Mv^2/2RT).dv.N

M = 16 x 10^-3 kg/mol

R = gas constant

N = Avogadro's number

dv = 500.001 - 500.000 = 0.001 m/s

v = 500.000 m/s

we get,

4piv^2 = (4 x 3.14 x (500)^2) = 3.14 x 10^6 m^2/s^2

(M/(2piRT)^3/2 = (16 x 10^-3/2 x 3.14 x 8.314 x 300)^3/2 = 1.03 x 10^-9 s^3/m^3

(Mv^2/2RT) = (16 x 10^-3 x (500)^2/2 x 8.314 x 300) = 0.802

Number of molecules in range = 3.14 x 10^6 x 1.02 x 10^-9 x e^(-0.802) x 0.001 x 6.023 x 10^23

= 8.64 x 10^17 molecules

(b) number of molecules = (4piv^2)(M/(2piRT)^3/2.(e^(-Mv^2/2RT).dv.N

M = 16 x 10^-3 kg/mol

R = gas constant

N = Avogadro's number

dv = 150.001 - 150.000 = 0.001 m/s

v = 150.000 m/s

we get,

4piv^2 = (4 x 3.14 x (150)^2) = 2.83 x 10^5 m^2/s^2

(M/(2piRT)^3/2 = (16 x 10^-3/2 x 3.14 x 8.314 x 300)^3/2 = 1.03 x 10^-9 s^3/m^3

(Mv^2/2RT) = (16 x 10^-3 x (150)^2/2 x 8.314 x 300) = 0.0722

Number of molecules in range = 2.83 x 10^5 x 1.02 x 10^-9 x e^(-0.07216) x 0.001 x 6.023 x 10^23

= 1.62 x 10^17 molecules

14.12 For CH4

number of molecules = (4piv^2)(M/(2piRT)^3/2.(e^(-Mv^2/2RT).dv.N

M = 16 x 10^-3 kg/mol

R = gas constant

N = Avogadro's number

dv = 400.001 - 400.000 = 0.001 m/s

v = 400.000 m/s

we get,

4piv^2 = (4 x 3.14 x (400)^2) = 2.01 x 10^6 m^2/s^2

(M/(2piRT)^3/2 = (16 x 10^-3/2 x 3.14 x 8.314 x 300)^3/2 = 1.03 x 10^-9 s^3/m^3

(Mv^2/2RT) = (16 x 10^-3 x (400)^2/2 x 8.314 x 300) = 0.5132

Number of molecules in range = 2.01 x 10^6 x 1.02 x 10^-9 x e^(-0.5132) x 0.001 x 6.023 x 10^23

= 7.40 x 10^17 molecules

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