Can You Find the Constant Value K for a Slowly Inflating Balloon?

In summary, the conversation is about a problem involving a spherical balloon that is being inflated and the equation used to describe its growth. The individual is looking for help in finding the value of the constant k, given a specific value for the radius and the rate of increase.
  • #1
Pin Head
23
0
Hi,
I'm working on a problem about a spherical balloon as its inflated slowly the radius of the balloon is in cm and the time is in seconds

dr/dt = k/r2

Now I know how to rearrange this equation using separable differentiation

r=3√3kt+c

However I would like to find the value of the constant K

For example the radius is 6cm and the radius is increasing at the rate of 5/36pi cm per second and I would like to rearrange the dr/dt=k/r2 to find the value of K

Any help would be appreciated
 
Physics news on Phys.org
  • #2
You are given dr/dt (5/36pi cm per second) at a given r (6cm). Plug these into your first equation and solve for k.
 

Related to Can You Find the Constant Value K for a Slowly Inflating Balloon?

What is the "constant value K" and why is it important?

The constant value K, also known as the equilibrium constant, is a measure of the ratio of products to reactants in a chemical reaction at equilibrium. It is important because it tells us the direction and extent of a reaction, as well as the concentrations of products and reactants at equilibrium.

How do you determine the value of K for a chemical reaction?

The value of K can be determined experimentally by measuring the concentrations of products and reactants at equilibrium and plugging them into the equilibrium expression. It can also be calculated using thermodynamic data or by using mathematical equations such as the van't Hoff equation.

What factors can affect the value of K?

The value of K can be affected by temperature, pressure, and the presence of a catalyst. Changes in these factors can shift the equilibrium position of a reaction, leading to a change in the value of K.

Why is the value of K sometimes different for the same reaction under different conditions?

The value of K can vary depending on the conditions under which the reaction is taking place. This is because factors such as temperature and pressure can affect the equilibrium position of the reaction, leading to a change in the concentrations of products and reactants and therefore a different value of K.

What does a high or low value of K indicate about a chemical reaction?

A high value of K indicates that the reaction favors the formation of products at equilibrium, while a low value of K indicates that the reaction favors the formation of reactants. Additionally, a very large or very small value of K can indicate a reaction that is proceeding either almost entirely in the forward or reverse direction, respectively.

Similar threads

I Real ODE yields real solution through complex numbers
    • Differential Equations
Replies
3
Views
2K
Calculations using the Standard Solar Model & Solar Equations of State
    • Advanced Physics Homework Help
Replies
1
Views
902
Find the derivative of a balloon's circumference
    • Calculus and Beyond Homework Help
Replies
4
Views
2K
I Find Practical Resonance Frequencies in Linear Differential Equations
    • Differential Equations
Replies
1
Views
2K
Differentiation with units- Related Rates problem
    • Calculus and Beyond Homework Help
Replies
30
Views
3K
How can the surface tension of a spherical balloon affect its radius over time?
    • Advanced Physics Homework Help
Replies
7
Views
1K
Engineering How do I find the differential equation for this circuit?
    • Engineering and Comp Sci Homework Help
Replies
28
Views
2K
Calculating time to reduce alcohol in wine using heating method
    • Chemistry
4
Replies
131
Views
4K
I How Do You Calculate the Dissolution Time of a Corroding Metal Sphere?
    • Differential Equations
Replies
3
Views
1K
Rates of change: surface area and volume of a sphere
    • Calculus and Beyond Homework Help
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top