Question

A tank contains 20 gallons of a solution, with 10 pounds of chemical \(\mathrm{A}\) in the solution. At a certain instant, we begin pouring in a solution containing the same chemical in a concentration of 2 pounds per gallon. We pour at a rate of 3 gallons per minute while simultaneously draining off the resulting (well-stirred) solution at the same rate. Find the amount of chemical A in the tank after 20 minutes.

Short answer

38.51 pounds of chemical A in the tank after 20 minutes.

Step-by-step solution

Set Up the Initial Condition

Initially, the tank contains 20 gallons of solution with 10 pounds of chemical A in it. Therefore, the initial concentration is \( \frac{10}{20} = 0.5 \) pounds per gallon.

Differential Equation for the Rate of Change

Let \( y(t) \) be the amount of chemical A in the tank at time \( t \). The rate at which chemical A enters the tank is \( 3 \times 2 = 6 \) pounds per minute since the incoming solution has a concentration of 2 pounds per gallon. The rate at which chemical A leaves the tank is \( 3 \times \frac{y(t)}{20} \), as the concentration in the tank is \( \frac{y(t)}{20} \) pounds per gallon. We can thus write the differential equation: \( \frac{dy}{dt} = 6 - \frac{3y(t)}{20} \).

Solve the Differential Equation

Rearrange the differential equation: \( \frac{dy}{dt} = 6 - \frac{3y}{20} \). This can be solved by separating variables or using an integrating factor. First, rearrange to a standard linear form: \( \frac{dy}{dt} + \frac{3y}{20} = 6 \). The integrating factor is \( e^{\int \frac{3}{20} dt} = e^{\frac{3}{20}t} \). Multiply through by this factor to get: \( e^{\frac{3}{20}t} \frac{dy}{dt} + \frac{3}{20} e^{\frac{3}{20}t} y = 6e^{\frac{3}{20}t} \). Integrate both sides to find \( y(t) \).

Integrating Both Sides

Integrate the left side with respect to \( t \): \( \int d\left( e^{\frac{3}{20}t} y \right) = \int 6e^{\frac{3}{20}t} dt \). The left integral is \( e^{\frac{3}{20}t} y \), and the right integral becomes \( \frac{6 \times 20}{3} e^{\frac{3}{20}t} = 40e^{\frac{3}{20}t} + C \). Thus, \( e^{\frac{3}{20}t} y = 40e^{\frac{3}{20}t} + C \).

Solve for the Constant

Use the initial condition \( y(0) = 10 \) to find \( C \): \( e^{0}10 = 40e^{0} + C \), so \( C = 10 - 40 = -30 \). Thus the solution is: \( e^{\frac{3}{20}t} y = 40e^{\frac{3}{20}t} - 30 \).

Solve for y(t)

Rearrange to express \( y(t) \): \( y(t) = 40 - 30e^{-\frac{3}{20}t} \).

Find the Amount of Chemical After 20 Minutes

Evaluate \( y(t) \) at \( t = 20 \): \( y(20) = 40 - 30e^{-3} \). Calculate \( e^{-3} \approx 0.0498 \); thus, \( y(20) = 40 - 30 \times 0.0498 \approx 40 - 1.494 \approx 38.51 \).

Conclusion

The amount of chemical A in the tank after 20 minutes is approximately 38.51 pounds.

Key concepts

Rate of Change

The concept of "rate of change" is crucial when dealing with differential equations, especially in problems involving dynamic systems like chemical mixtures. In this context, the rate of change describes how the quantity of a chemical in a solution is changing over time.
The rate of a process is often given in terms of time, for example, pounds of chemical per minute as in this exercise.
To analyze the rate of change of chemical A in the tank, we first determine how it enters and exits the system.
- **Entering Rate**: Here, the chemical enters the tank at a rate of 6 pounds per minute because you are adding a solution with 2 pounds of chemical per gallon at 3 gallons per minute.- **Exiting Rate**: The exiting rate depends on the concentration, which is variable over time. This rate is computed as (3 gallons/minute) times the concentration at any time.So, the differential equation captures how these rates balance: \( \frac{dy}{dt} = \, \text{(entering rate)} - \, \text{(exiting rate)} \).
This results in: \( \frac{dy}{dt} = 6 - \frac{3y(t)}{20} \), showing how chemical A changes over time.

Concentration in Solution

In problems related to solutions and mixtures, concentration is a critical factor.
It determines how much of a substance is present in a specific volume of solution. For this exercise:- Initially, the tank contains 10 pounds of chemical A in 20 gallons, leading to a concentration of \( \frac{10}{20} = 0.5 \) pounds per gallon.- As time progresses, this concentration changes due to the continual in-out flow of liquid.
Understanding concentration involves realizing that with every change in the fluid's amount (from adding in or draining), the proportion or concentration of the chemical alters too.
During the process, the incoming solution has a fixed concentration of 2 pounds per gallon, which affects the overall concentration in the tank. Hence, the changing concentration over time must be considered integrally while solving the equation.

Integrating Factor Method

The Integrating Factor Method is a powerful tool used for solving linear first-order differential equations.
These kinds of equations appear in many problems involving rates, such as mixing processes.In the original exercise, the differential equation is rearranged into the form:\( \frac{dy}{dt} + \frac{3y}{20} = 6 \), and we identify the integrating factor as:\[ e^{\int \frac{3}{20} \, dt} = e^{\frac{3}{20}t} \].
Applying the integrating factor transforms the left side into the derivative of:\( e^{\frac{3}{20}t}y(t) \), simplifying the equation to:\( \frac{d}{dt}(e^{\frac{3}{20}t}y) = 6e^{\frac{3}{20}t} \).
We integrate both sides with respect to \( t \) to find the function \( y(t) \).
The solution involves an integration constant, which we determine using the initial condition. Essentially, the integrating factor helps convert a complex differential scenario into a solvable integral problem.

Chemical Mixture Problem

Chemical mixture problems often involve concepts from chemistry and differential equations.
These problems ask us to determine how a certain substance's quantity changes in a mixture over time. The key challenge in these problems is that substances continuously mix, enter, and exit the solution, affecting the concentration.
The given exercise exemplifies a typical chemical mixture scenario with fixed rates of input and output. - Here, chemical A enters at a constant rate while its concentration continuously changes inside the solution due to mixing. - Simultaneously, an equal volume exits, ensuring a constant total volume. To solve it, a differential equation models the balance of incoming and outgoing chemicals. This allows us to predict changes in concentration over time and determines characteristics like final concentration after a specified duration.
As a result, mastering these problems involves setting up the correct differential equation and applying methods like separation of variables or the integrating factor to find the solution.