# word problem

• Apr 1st 2008, 08:26 AM
word problem
The volume V of a gas varies inversely as the pressure P upon it. The volume of a gas is 200 cu cm under a pressure of 32kg/sq. cm
a. find the variation constant
b. what will be its volume under a presure of 40 kg/sq. cm
• Apr 1st 2008, 08:54 AM
Jhevon
Quote:

The volume V of a gas varies inversely as the pressure P upon it. The volume of a gas is 200 cu cm under a pressure of 32kg/sq. cm
a. find the variation constant
b. what will be its volume under a presure of 40 kg/sq. cm

Hint: to vary inversly means they have this relationship.

$\displaystyle V = \frac kT$ .........for $\displaystyle k$ some constant (call the proportionality constant)
• Apr 1st 2008, 09:17 AM
im sorry that made no sense to me lol i am so lost
• Apr 1st 2008, 09:23 AM
janvdl
Quote:

The volume V of a gas varies inversely as the pressure P upon it. The volume of a gas is 200 cu cm under a pressure of 32kg/sq. cm
a. find the variation constant
b. what will be its volume under a presure of 40 kg/sq. cm

I think Jhevon should have wrote it as:

$\displaystyle V = \frac{k}{P}$

Where $\displaystyle V$ is the volume and $\displaystyle P$ the pressure.

In the first question, to find the constant $\displaystyle k$ you need to put the given values for $\displaystyle V$ and $\displaystyle P$ in the equation, and solve for $\displaystyle k$.

In the second question you now make use of that value for $\displaystyle k$ when solving for the volume, $\displaystyle V$.
• Apr 1st 2008, 09:40 AM
Jhevon
Quote:

Originally Posted by janvdl
I think Jhevon should have wrote it as:

$\displaystyle V = \frac{k}{P}$

Where $\displaystyle V$ is the volume and $\displaystyle P$ the pressure.

In the first question, to find the constant $\displaystyle k$ you need to put the given values for $\displaystyle V$ and $\displaystyle P$ in the equation, and solve for $\displaystyle k$.

In the second question you know make use of that value for $\displaystyle k$ when solving for the volume, $\displaystyle V$.

yes, i meant P not T :D
• Apr 1st 2008, 12:27 PM
i am confused how to solve for k
• Apr 1st 2008, 12:55 PM
Ting
As janvdl said, substitute in the given values of V and P and solve for k.

$\displaystyle 200 = \frac{k}{32}$

Solve for k, multiply both sides by 32.
• Apr 1st 2008, 01:31 PM
32*200=6400 is that the value of k
• Apr 1st 2008, 01:33 PM
janvdl
Quote:

Yes. Now use that value for $\displaystyle k$ and the given value for $\displaystyle P$, to solve for $\displaystyle V$. (The second question.)
$\displaystyle V = \frac{6400}{P}$