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2 years ago

9

A river with a flow rate of 7 m3/s discharges into a lake with a volume of 9,000,000 m3. The concentration of a particular VOC i

n the river is 3 mg/L. The lake has a decay coefficient for the VOC of 0.3 per day. What is the concentration of the VOC in the river downstream?

1 answer:

soldi70 [24.7K]2 years ago

4 0

Answer:

The concentration downstream reduces to 0.03463mg/L

Explanation:

Initially let us calculate the time it is required to fill the lake, since for that period of time the pollutant shall remain in the lake before being flushed out.

Thus the detention period is calculated as

$t=\frac{V}{Q}\\\\t=\frac{9000000}{7}=1.285\times 10^{6}seconds\\\\\therefore t = 14.872days$

Now the concentration of the pollutant after 14.872 days is calculated as

$N_{t}=N_{0}e^{-kt}$

where

$N_{o}$ is the initial concentration

't' is the time elapsed after which the remaining concentration is calculated

k is the dissociation constant.

Applying values we get

$N_{t}=3\times e^{-0.3\times 14.872}$

$N_{t}=0.03463mg/L$

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Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(c) $v_{1} = 2.04\times 10^{8}\ m/s$

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Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

(c) To calculate the velocity of light in medium 1:

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

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