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Chapter 13: Q68P (page 359)

(II) Estimate the time needed for a glycine molecule (see Table 13鈥�4) to diffuse a distance of \(25\;\mu {\rm{m}}\) in water at 20掳C if its concentration varies over that distance from \(1.00\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}\) to \(0.50\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}\)? Compare this 鈥渟peed鈥� to its rms (thermal) speed. The molecular mass of glycine is about 75 u.

Short Answer

Expert verified

The time for the diffusion is 0.99 s.

The rms speed is larger than the diffusion speed.

Step by step solution

01

Given data

The higher concentration is \({C_1} = 1.00\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}\).

The lower concentration is \({C_2} = 0.50\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}\).

The distance is \(\Delta x = 25\;\mu {\rm{m}} = 25 \times {10^{ - 6}}\;{\rm{m}}\).

The temperature is \(T = {20^ \circ }{\rm{C}} = 293\;{\rm{K}}\).

The molar mass of the glycine is \(m = 75\;u = 75 \times 1.66 \times {10^{ - 27}}\;{\rm{kg}}\).

The diffusivity of the glycine is \(D = 95 \times {10^{ - 11}}\;{{\rm{m}}^{\rm{2}}}{\rm{/s}}\).

02

Concepts

The rms thermal speed of any gas is\({v_{{\rm{rms}}}} = \sqrt {\frac{{3kT}}{m}} \).

The time for the diffusion is\(t = \frac{{\bar C}}{{\Delta C}}\frac{{{{\left( {\Delta x} \right)}^2}}}{D}\).

03

Calculation of the diffusion time

The average concentration is \(\bar C = \frac{1}{2}\left( {{C_1} + {C_2}} \right)\).

The difference between the concentrations is \(\Delta C = {C_1} - {C_2}\).

The time for the diffusion is:

\(\begin{aligned}{c}t &= \frac{{\bar C}}{{\Delta C}}\frac{{{{\left( {\Delta x} \right)}^2}}}{D}\\t &= \frac{{\frac{1}{2}\left( {{C_1} + {C_2}} \right)}}{{{C_1} - {C_2}}}\frac{{{{\left( {\Delta x} \right)}^2}}}{D}\\t &= \frac{{\frac{1}{2}\left( {1.00\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}} + 0.50\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}} \right)}}{{1.00\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}} - 0.50\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}}}\frac{{{{\left( {25 \times {{10}^{ - 6}}\;{\rm{m}}} \right)}^2}}}{{95 \times {{10}^{ - 11}}\;{{\rm{m}}^{\rm{2}}}{\rm{/s}}}}\\t &= 0.99\;{\rm{s}}\\\end{aligned}\)

Hence, the required time for the diffusion is 0.99 s.

04

Comparison between diffusion speed and thermal speed

The diffusion speed is:

\(\begin{aligned}{c}{v_d} &= \frac{{\Delta x}}{t}\\ &= \frac{{25 \times {{10}^{ - 6}}\;{\rm{m}}}}{{0.99\;{\rm{s}}}}\\ &= 2.5 \times {10^{ - 5}}\;{\rm{m/s}}\end{aligned}\)

The rms speed of the glycine molecules is:

\(\begin{aligned}{c}{v_{{\rm{rms}}}} &= \sqrt {\frac{{3kT}}{m}} \\ &= \sqrt {\frac{{3 \times \left( {1.38 \times {{10}^{ - 34}}\;{\rm{J/K}}} \right) \times \left( {293\;{\rm{K}}} \right)}}{{75 \times 1.66 \times {{10}^{ - 27}}\;{\rm{kg}}}}} \\ &= 312\;{\rm{m/s}}\end{aligned}\)

Therefore, the rms speed is much greater than the diffusion speed.

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Most popular questions from this chapter

In outer space, the density of matter is about one atom per cm3, mainly hydrogen atoms, and the temperature is about 2.7 K. Calculate the RMSspeed of these hydrogen atoms and the pressure (in atmospheres).

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Question: (II)What is the rms speed of nitrogen molecules contained in an\({\bf{8}}{\bf{.5}}\;{{\bf{m}}^{\bf{3}}}\)volume at 2.9 atm if the total amount of nitrogen is 2100 mol?

The escape speed from the Earth is \({\bf{1}}{\bf{.12 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{m/s}}\),that is, a gas molecule traveling away from Earth near the outer boundary of the Earth鈥檚 atmosphere would, at this speed, be able to escape from the Earth鈥檚 gravitational field and be lost in the atmosphere. At what temperature is the RMS speed of (a) oxygen molecules and (b) helium atoms equal to \({\bf{1}}{\bf{.12 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{m/s}}\)? (c) Can you explain why our atmosphere contains oxygen but not helium?

Question:(II) A certain car has 14.0 L of liquid coolant circulating at a temperature of 93掳C through the engine鈥檚 cooling system. Assume that, in this normal condition, the coolant completely fills the 3.5-L volume of the aluminum radiator and the 10.5-L internal cavities within the aluminum engine. When a car overheats, the radiator, engine, and coolant expand and a small reservoir connected to the radiator catches any resultant coolant overflow. Estimate how much coolant overflows to the reservoir if the system goes from 93掳C to 105掳C. Model the radiator and engine as hollow shells of aluminum. The coefficient of volume expansion for coolant is\({\bf{410}} \times {\bf{1}}{{\bf{0}}^{{\bf{ - 6}}}}\;{\bf{/^\circ C}}\).
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