PLease give me the equation

Question

A certain chemical dissolves in water at a rate proportional to the product of the amount
undissolved and the difference between the concentration in a saturated solution and
the concentration in the actual solution. In 100 grams of a saturated solution it is known
that 50 grams of the substance are dissolved. If when 30 grams of the chemical are
agitated with 100 grams of water, 10 grams are dissolved in 2 hours, how much will be
dissolved in 5 hours?

Answer 1

This is a differential equations problem involving the rate at which a chemical dissolves in water.

We will use the information given to create the differential equation and then determine how much chemical dissolves after 5 hours.

Step 1: Define the Variable

Let x(t) = the number of grams of chemical dissolved after t hours.

Initially, none of the chemical has dissolved:

x(0) = 0

Since we start with 30 grams of chemical:

Amount undissolved = 30 - x

Step 2: Determine the Concentrations

We are told that 100 grams of saturated solution contains 50 grams of dissolved chemical.

Therefore, the saturated concentration is:

Cs = 50/100 = 0.5

The actual solution contains 100 grams of water and x grams of dissolved chemical.

Total mass of solution = 100 + x

Therefore, the actual concentration is:

C = x/(100 + x)

The difference between the saturated and actual concentrations is:

0.5 - x/(100 + x)

Step 3: Set Up the Differential Equation

The problem states that the rate of dissolving is proportional to:

1. The amount of chemical remaining undissolved.
2. The difference between the saturated concentration and actual concentration.

Therefore:

dx/dt = k(30 - x)(0.5 - x/(100 + x))

Where k is a positive proportionality constant.

This is the differential equation requested.

We can simplify it:

0.5 - x/(100 + x) = (100 - x)/(2(100 + x))

Therefore:

dx/dt = k(30 - x)(100 - x)/[2(100 + x)]

Step 4: Use the Information Given

We know that 10 grams dissolve in 2 hours.

Therefore:

x(2) = 10

Separate the variables:

2(100 + x)/[(30 - x)(100 - x)] dx = k dt

Using partial fractions:

2(100 + x)/[(30 - x)(100 - x)]

= (26/7)/(30 - x) - (40/7)/(100 - x)

Integrating gives:

-(26/7)ln(30 - x) + (40/7)ln(100 - x) = kt + C

Using x(0) = 0:

(26/7)ln[30/(30 - x)] + (40/7)ln[(100 - x)/100] = kt

Now substitute x = 10 and t = 2:

(26/7)ln(30/20) + (40/7)ln(90/100) = 2k

Solving:

k ≈ 0.45198

Step 5: Find the Amount Dissolved After 5 Hours

Substitute t = 5:

(26/7)ln[30/(30 - x)] + (40/7)ln[(100 - x)/100] = 5(0.45198)

Solving numerically:

x ≈ 17.96 grams

Final Answer:

The differential equation is:

dx/dt = k(30 - x)(0.5 - x/(100 + x))

Using the initial conditions, approximately 17.96 grams of the chemical will be dissolved after 5 hours.

Key Takeaway:

The rate of dissolving decreases as more chemical dissolves because the amount remaining undissolved becomes smaller and the solution's concentration increases.

This solution assumes concentration means the mass of dissolved chemical divided by the total mass of the solution.