P is a fixed mass of gas at constant temperature is inversely proportional to its volume, V cubed. When v=2, p=125.

A) Find the value of V when P=325

B) Find the value of P when V=10

Guest Aug 21, 2017

edited by
Guest
Aug 21, 2017

#1**+1 **

From the question we can deduct that the temperature is proportional to the volume of the gas, set:

\(P=\frac{C}{V}\)

With P being pressure, V being the volume, and C being the constant of the gas.

A. Plug \(V=2\) and \(P=125\) into the equation:

\(125=\frac{C}{2}\)

\(C=250\)

Therefore the relationship between \(P\) and \(V\) is:

\(P=\frac{250}{V}\)

Plug \(P=325\) into the equation:

\(325=\frac{250}{V}\)

Multiply both sides of the equation by a factor of \(V\):

\(325V=250\)

Divide both sides by \(325\):

\(V=\frac{250}{325}=\frac{10}{13}\) (Unit Volume)

Done :D

B.

Plug \(V=10\) into the equation:

\(P=\frac{250}{10}=25\) (Unit Pressure)

Done :D x2

Jeffes02
Aug 21, 2017

#4**+1 **

The question does not make sense to me. How can the mass of an object be inversely proportional to its volume, much less its volume cubed!

Then again the question calls it a fixed mass, if it is a fixed mass then at a set temperature it would have a fixed voume. There are no variables, everything is fixed.

Overlooking all these 'problems'

\(P=\frac{k}{V^3}\\ 125=\frac{k}{8}\\ k=1000\\ P=\frac{1000}{V^3}\\ A) \\When\;\; P=325\\ 325V^3=1000\\ V=\frac{10}{\sqrt[3]{325}}\\ V\approx 1.454\\~\\ B)\\When\;\;V=10\\ P=\frac{1000}{20^3}=1\)

Melody
Aug 21, 2017

#5**0 **

Oops I forgot to put in that P was Pascals (pressure), I hope that makes more sense now

Guest Aug 21, 2017

#8**+3 **

Use Boyle's Law

For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure. (The written form introduces ambiguities. Volume is measured in cubic units. It is not the cube of the volume.)

Boyle's Law expressed mathematically:

\(P_1 V_1 = P_2 V_2\\ \text{ }\\ \text{1)}\\ V_2 = \dfrac {P_1 \cdot V_1}{P_2}\\ V_2 = \dfrac {125\; units \cdot 2\;units^3}{325\; units} = 0.76923\; units^3\\ \)

\(\text{2)}\\ P_2 = \dfrac {P_1 \cdot V_1}{V_2}\\ P_2 = \dfrac {125\; units \cdot 2\;units^3}{10\; units} = 25\; mass \;units\; per \;area \;units^2 \\ \)

GingerAle
Aug 21, 2017