Integration Help: Solve d/dt(Vc1) = Ak[c1(sat) - c1]

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In summary, the problem is trying to find the concentration of a gas that is in equilibrium with another gas. The equation has unknowns, including the concentration of the first gas and the pressure of the second gas. The differential equation was solved to find the concentration of the first gas, and then the second gas was solved for to find the pressure.
  • #1
Triathlete
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Homework Statement



I am working on a mass transfer problem and have this equation:

d/dt(Vc1) = Ak[c1(sat) - c1]

Homework Equations



Initial conditions:

x=0, c1=0

The Attempt at a Solution



I know that the result from integration should be:

c1/c1(sat) = 1 – e-(kA/V)t

But I don't understand the integration that got there. If someone could show me the steps, that would be very helpful!
Thanks!
 
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  • #2
Triathlete said:

Homework Statement



I am working on a mass transfer problem and have this equation:

d/dt(Vc1) = Ak[c1(sat) - c1]
What are the symbols here? What's the difference between c1 and c1(sat). Does Vc1 mean V * c1 or is it ##V_{c_1}##
Triathlete said:

Homework Equations



Initial conditions:

x=0, c1=0

The Attempt at a Solution



I know that the result from integration should be:

c1/c1(sat) = 1 – e-(kA/V)t

But I don't understand the integration that got there. If someone could show me the steps, that would be very helpful!
Thanks!
 
  • #3
Thanks for the response, c1 is the concentration c1(sat) is the saturated concentration. It is supposed to be V*c1.
 
  • #4
## \frac{d}{dt}Vc_1 =Ak [ c_1 (sat) - c_1 ]##
This problem is easier to do as a differential equation.

Let's call this ##V c'(t) = Ak c_{sat} - Akc(t) ##.
This gives:
##V c'(t) +Akc(t) = Ak c_{sat}\\
c'(t) + \frac{Ak}{V} c(t) =\frac{Ak}{V} c_{sat} ##
A general solution to
##c'(t) + \frac{Ak}{V} c(t)=0## is ##c(t) = N e^{- \frac{Ak}{V} t} ##.
Where N is a constant.
Using this general solution, you can solve for a particular solution that satisfies the differential form you were given and initial conditions.
Notice that you can just add ##c_{sat}## to ##c(t) ## without affecting the derivative, since it is a constant.
So a solution to
##c'(t) + \frac{Ak}{V} c(t)= \frac{Ak}{V}c_{sat}## is ##c(t) = N e^{- \frac{Ak}{V} t} +c_{sat} ##.
Now, use your initial condition to solve for N.
##c(t) = N e^{- \frac{Ak}{V} t} +c_{sat} ## with ##c(0) = 0##.
...
From there, you should see how the solution came about.
 
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Related to Integration Help: Solve d/dt(Vc1) = Ak[c1(sat) - c1]

1. What does the equation d/dt(Vc1) = Ak[c1(sat) - c1] represent?

The equation represents the rate of change of voltage (Vc1) over time, which is equal to the product of the rate constant (A), the difference between the saturation concentration (c1(sat)) and the actual concentration (c1).

2. How is this integration helpful in solving problems?

This integration allows us to model and understand the behavior of a system over time. By solving for the concentration (c1) at different time points, we can gain insight into the kinetics and dynamics of the system.

3. What are the steps involved in solving this type of integration problem?

The first step is to rearrange the equation and separate the variables, in this case, c1 and t. Then, we integrate both sides and solve for c1. Finally, we plug in the known values for Vc1, A, and c1(sat) to obtain the final solution.

4. Can this equation be applied to any system?

Yes, this equation can be applied to any system where the concentration of a substance (c1) is changing over time due to a reaction with a rate constant (A) and a saturation concentration (c1(sat)). However, the specific values for Vc1, A, and c1(sat) will vary depending on the system being studied.

5. How does the saturation concentration (c1(sat)) affect the rate of change of Vc1?

The saturation concentration is a measure of the maximum amount of substance that can be present in the system. As the actual concentration (c1) approaches the saturation concentration, the rate of change of Vc1 decreases, eventually reaching a steady state where Vc1 remains constant. This is because the reaction rate is limited by the availability of the reactant (c1) at higher concentrations.

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