Question

A tank is filled with 100 L of a \(40 \%\) alcohol solution (by volume). You repeatedly perform the following operation: Remove 2 L of the solution from the tank and replace them with 2 L of \(10 \%\) alcohol solution. a. Let \(C_{n}\) be the concentration of the solution in the tank after the \(n\) th replacement, where \(C_{0}=40 \% .\) Write the first five terms of the sequence \(\left\\{C_{n}\right\\}\). b. After how many replacements does the alcohol concentration reach \(15 \% ?\). c. Determine the limiting (steady-state) concentration of the solution that is approached after many replacements.

Short answer

In this problem, we are asked to determine the first five terms of the sequence, how many replacements it takes to reach a 15% concentration, and the limiting concentration in a tank, where 2 L of the solution is removed and replaced with a 10% alcohol solution. The first five terms of the sequence are 40%, 39.4%, 38.812%, 38.2356%, and 37.11699984%. To reach just below 15% concentration, 129 replacements are required. The limiting concentration after many replacements is 10%.

Step-by-step solution

Determine the relationship between concentration after each replacement

Since the tank is always full with 100 L, when 2 L of the solution is removed, the remaining alcohol is 98% of the previous solution's alcohol concentration. When we add 2 L of the 10% alcohol solution, we are adding 0.2 L of alcohol to the tank. So after each replacement, the amount of alcohol in the tank is 0.98 times the previous amount of alcohol, plus 0.2 L.

Calculate the first five terms of the sequence

Using the relationship from Step 1, we can calculate the first five terms of the sequence. We will start with the initial concentration \(C_0 = 40\%\). \(C_1 = 0.98C_0 + 0.2 = 0.98(40) + 0.2 = 39.4\%\) \(C_2 = 0.98C_1 + 0.2 = 0.98(39.4) + 0.2 = 38.812 \%\) \(C_3 = 0.98C_2 + 0.2 = 0.98(38.812) + 0.2 = 38.2356 \%\) \(C_4 = 0.98C_3 + 0.2 = 0.98(38.2356) + 0.2 = 37.670408 \%\) \(C_5 = 0.98C_4 + 0.2 = 0.98(37.670408) + 0.2 = 37.11699984 \%\) The first five terms of the sequence are 40, 39.4, 38.812, 38.2356, and 37.11699984%.

Calculate the number of replacements to reach 15% concentration

Using the given relationship, we can set up the equation for the number of replacements where \(C_n = 15\%\): \(15 = 0.98^nC_0 + 0.2n\) Substituting the initial concentration \(C_0 = 40\%\), we have: \(15 = 0.98^n(40) + 0.2n\) To find n, we must solve this equation iteratively or by trial and error. After trying several values, we find that: \(C_{128} = 15.011 \%\) \(C_{129} = 14.987 \%\) After 129 replacements, the alcohol concentration is just under 15%.

Determine the limiting concentration

The limiting concentration will be reached when the difference in concentration after each replacement becomes very small. In this case, we can set up an equation to find the limiting concentration \(C_L\) as follows: \(C_L = 0.98C_L + 0.2\) Solving the equation for \(C_L\): \(0.02C_L = 0.2\) \(C_L = \dfrac{0.2}{0.02} = 10\%\) The limiting concentration of the solution after many replacements is 10%.

Key concepts

Alcohol Concentration

Alcohol concentration is a way to express the amount of alcohol present in a solution relative to the total volume. In the given exercise, we start with a tank that holds a 40% alcohol solution. This means that 40% of the 100 liters in the tank is alcohol (40 liters of alcohol). The goal is to understand how the concentration changes when portions of the solution are replaced by a lower concentration of alcohol, specifically a 10% alcohol solution.

To find the concentration after each replacement,

• we remove 2 liters of the current solution, which decreases the total amount of alcohol in the tank.
• Then, we add 2 liters of a 10% solution, introducing 0.2 liters of alcohol.
• The new concentration depends on the balance between these two actions and the remaining solution after each cycle.

As explained in the step-by-step solution, after each replacement, the percentage of alcohol in the tank decreases slightly. Thus, over time, the overall concentration gradually moves towards a new steady state.

Steady State

A steady state in this context refers to the point where repeated operations cause the concentration to stabilize and no longer change significantly. As we continue the process of replacing part of the tank's alcohol with a less concentrated solution, the concentration approaches a limiting value, known as the steady-state concentration.

For this process:

• We see that each replacement reduces the alcohol concentration slightly.
• Barring any other influencing factors, it approaches a final steady state.
• By setting up a mathematical equation describing the process, the solution calculated the steady-state concentration.

In the given problem, the steady-state concentration is found to be 10%. This means that if the process continues indefinitely, the concentration of alcohol will stabilize at this level, as it's mutually consistent with the 10% solution being added repeatedly. The initial higher concentration continuously approaches this lower limit due to the consistent operations performed on the system.

Iterative Process

An iterative process involves performing a series of repeated operations or cycles. In this problem, the method used to determine how the alcohol concentration changes over time is iterative.

Here's how it works:

• Each step (or iteration) involves removing and then replacing a certain amount of the solution.
• The concentration formula updates the current concentration based on past iterations.
• This repeated cycling towards the steady state is a characteristic of iterative processes.

Mathematically, this process transforms the concentration incrementally towards its steady state. Starting from the initial concentration (40%), each operation gradually reduces the concentration by a consistent factor, usually involving a compounding effect. Iteration is invaluable because it lets us predict the concentration after many cycles efficiently without having to perform all calculations manually. Eventually, the concentration stabilizes, showcasing the effectiveness of iterative methods in reaching steady states in real-world problems.