word problem

**imbadatmath** · April 1st 2008, 08:26 AM

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

**Jhevon** · April 1st 2008, 08:54 AM

Originally Posted by imbadatmath

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.

$V = \frac kT$ .........for $k$ some constant (call the proportionality constant)

**imbadatmath** · April 1st 2008, 09:17 AM

im sorry that made no sense to me lol i am so lost

**janvdl** · April 1st 2008, 09:23 AM

Originally Posted by imbadatmath

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:

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

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

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

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

**Jhevon** · April 1st 2008, 09:40 AM

Originally Posted by janvdl

I think Jhevon should have wrote it as:

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

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

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

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

yes, i meant P not T

**imbadatmath** · April 1st 2008, 12:27 PM

i am confused how to solve for k

**Ting** · April 1st 2008, 12:55 PM

As janvdl said, substitute in the given values of V and P and solve for k.

$<br /> 200 = \frac{k}{32}<br />$

Solve for k, multiply both sides by 32.

**imbadatmath** · April 1st 2008, 01:31 PM

32*200=6400 is that the value of k

**janvdl** · April 1st 2008, 01:33 PM

Originally Posted by imbadatmath

32*200=6400 is that the value of k

Yes. Now use that value for $k$ and the given value for $P$ , to solve for $V$ . (The second question.)

**Ting** · April 1st 2008, 01:39 PM

Yes so you have the proportionality formula:

$<br /> V = \frac{6400}{P}<br />$

To find V when P = 40, substutute P = 40 into the formula.

**imbadatmath** · April 1st 2008, 02:16 PM

so the volume in the second one owuld equal 160

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