
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 30th, 2017, 08:24 PM  #1 
Senior Member Joined: Jan 2017 From: US Posts: 111 Thanks: 6  Solve For Pressure Applied The volume V of a gas varies inversely as the pressure P on it. If the volume is 250 cm$\displaystyle ^3$ under a pressure of 35 kg/cm$\displaystyle ^3$, solve for the pressure applied to have a volume of 180 cm$\displaystyle ^3$. Be sure to define your variables, find your constant variable, construct your variation equation, and then solve. 
May 2nd, 2017, 01:23 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,115 Thanks: 708 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
The question states the steps you need to take to solve it. 1. Define your variables 2. Find your constant variable 3. Construct your variation equation 4. Solve it The question also hints at the things that are going to be variables in your equation, volume and pressure, and how they are related to each other. Therefore, have a go and let us know how it goes 
May 3rd, 2017, 07:05 PM  #3 
Senior Member Joined: Jan 2017 From: US Posts: 111 Thanks: 6  V= 250 cm$\displaystyle ^3$ when P=35 kg/cm$\displaystyle ^3$ Correct me if I'm wrong, but since the volume depends on the pressure applied, the constant variable would be V=250cm$\displaystyle ^3$, right? So: y=250cm$\displaystyle ^3$ y=kx when p=35 kg/cm$\displaystyle ^3$, v=250cm$\displaystyle ^3$ 35kg/cm$\displaystyle ^3$=k*250cm$\displaystyle ^3$ So the equation would be 35kg/cm$\displaystyle ^3$=250cm$\displaystyle ^3$*x. Am I at all on the right track here? 
May 4th, 2017, 02:10 AM  #4  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,115 Thanks: 708 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
Here's some further information that might help: There are different names for different quantities in equations depending on their behaviour.  Variables: These are quantities that can change and are usually things that describe the state of a system at a particular point in time.  Independent variables: These are particular variables that do not depend on anything. Usually they are things like space and time, which you characterise everything else in terms of. Most calculations are decribed as 1D, 2D, 3D, etc. based on the number of independent variables used (hence some people describing 4D as referring to 3 space variables and 1 time variable).  Dependent variables: These are particular variables which have dependencies on other things. They have relationships between each other which describe how they vary. The relationships can be described using proportionalities or functions/formulae.  Constant: These are quantities that do not change. Therefore, they usually describe the system as a whole or how its processes are working. Variables that just happen to stay the same for some sort of process are sometimes called constants.  Parameters: These are constants that are required to solve something at the beginning of a calculation. They are often called input parameters. Now some information on relationships between variables. Consider two variables, $\displaystyle a$ and $\displaystyle b$...  Proportional This means that if you increase one thing, you increase the other (so if you double one quantity, you double the other). Written down, this is $\displaystyle a \propto b$ where the $\displaystyle \propto$ symbol just means "proportional to". You can turn any proportionality into an equation by multiplying the righthand side by a constant (most people use k) and swapping the $\displaystyle \propto$ with an equals sign. So you end up with $\displaystyle a = kb$ If you plot $\displaystyle a$ as a function of $\displaystyle b$ you get a straight line on a plot going through the origin, so this relationship is often called a "linear relationship".  Inversely proportional This means that if you increase one thing, you decrease the other (so if you double one quantity, you halve the other). Written down, this is $\displaystyle a \propto \frac{1}{b}$ Making an equation from this looks like $\displaystyle a = \frac{k}{b}$ Now onto your gas problem... In the question there is a statement: "volume of a gas varies inversely as the pressure". That means that both the volume and the pressure are variables and the relationship between those two variables is inverse proportionality (i.e. you halve one thing, you double the other). Written down, this looks like $\displaystyle V \propto \frac{1}{P}$ Turning this into an equation gives $\displaystyle V = \frac{k}{P}$ where k is the constant. You can then substitute the initial volume and pressure parameters to find k. Once you have k, you can use that together with the target volume to get the pressure required. Last edited by Benit13; May 4th, 2017 at 02:15 AM.  

Tags 
applied, pressure, solve 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
partial air pressure  srecko  Chemistry  1  January 8th, 2017 07:02 AM 
Pressure equation  Studentno666  Physics  3  July 13th, 2015 02:17 AM 
fluid pressure  hobilla  PreCalculus  4  March 21st, 2015 05:11 AM 
The lowest pressure  Phatossi  Physics  1  June 17th, 2012 12:29 PM 