Question

according to our textbook, at 25°c, the diffusion constant of co₂ in water is d_water = 1.9×10⁻⁹ m²/s and the diffusion constant of co₂ in air at one atmosphere is d_air = 2×10⁻⁵ m²/s. if it takes t_air = 4 s for co₂ to diffuse a distance r_rms = d_1 through the air, how long does it take for co₂ to diffuse r_rms = d_1 through the water? part 1 prepare: write a formula for the diffusion time t in terms of the distance d_1 and diffusion constant d given that r_rms = √(4dt). note that it may not be necessary to use every variable. use the following table as a reference for using each variable: d_1 d1 d d t = ((d1)^2/(4*d))

Explanation:

Step1: Rearrange the diffusion - time formula

We are given the formula $r_{ms}=\sqrt{4Dt}$. Squaring both sides gives $r_{ms}^2 = 4Dt$. Then, we can solve for $t$: $t=\frac{r_{ms}^2}{4D}$.

Step2: Substitute the values for water

We know that for water, when $r_{ms}=d_1$ and $D = D_{water}=1.9\times10^{-9}\ m^2/s$. Also, we know from the air - case that when $D = D_{air}=2\times10^{-5}\ m^2/s$, $t_{air}=4\ s$ and $r_{ms}=d_1$. First, from the air - case $d_1^2=4D_{air}t_{air}$. Then, for the water - case, $t_{water}=\frac{d_1^2}{4D_{water}}$. Substituting $d_1^2 = 4D_{air}t_{air}$ into the water - time formula, we get $t_{water}=\frac{4D_{air}t_{air}}{4D_{water}}=\frac{D_{air}t_{air}}{D_{water}}$.

Step3: Calculate the time in water

Substitute $D_{air}=2\times10^{-5}\ m^2/s$, $t_{air}=4\ s$, and $D_{water}=1.9\times10^{-9}\ m^2/s$ into the formula $t_{water}=\frac{D_{air}t_{air}}{D_{water}}$.
\[t_{water}=\frac{2\times 10^{-5}\ m^2/s\times4\ s}{1.9\times10^{-9}\ m^2/s}\]
\[t_{water}=\frac{8\times10^{-5}}{1.9\times10^{-9}}\ s\]
\[t_{water}=\frac{8}{1.9}\times10^{4}\ s\approx4.21\times 10^{4}\ s\]

Answer:

$t_{water}\approx4.21\times 10^{4}\ s$