11.3: Boyle's Law
- Page ID
- 116669
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Boyle's Law and Its Significance
The effect of compressing air on its pressure is governed by Boyle's Law, a fundamental principle that describes the relationship between pressure and volume in a gas at a constant temperature. Boyle's Law states:
P1×V1=P2×V2
Where:
P1 is the initial absolute pressure
V1 is the initial volume
P2 is the final absolute pressure
V2 is the final volume
Boyle’s Law reveals that pressure and volume are inversely proportional—if the volume is reduced, pressure increases proportionately, provided temperature remains constant. For instance, if the air volume is halved, the pressure will double. This law provides an essential calculation tool in pneumatic design, allowing for precise adjustments in component sizing, such as storage tanks. Properly sizing the storage tank is crucial for ensuring that compressors operate efficiently and don’t overheat due to constant use.
Suppose we have a sample of gas initially at a gage pressure of 30 PSI (PSIg) and a volume of 4 cubic feet (ft³). If the volume of the gas is decreased to 2 cubic feet while the temperature remains constant, what will be the new pressure of the gas?
We can still use Boyle's Law equation P1×V1=P2×V2 to solve this problem. Here's the solution in terms of cubic feet:
Step 1: Identify Initial and Final Conditions:
Initial pressure (P1): 30 PSIg
(This will need to be converted to Atmospheric Gage reading by adding 14.7 psi.)
(P1): 30 PSIg + 14.7 psi = 44.7 PSIa
Initial volume (V1): 4 ft³
Final volume (V2): 2 ft³ (since the volume decreases)
Step 2: Set Up the Equation: Using Boyle's Law equation: P1×V1=P2×V2
Given: P1×V1=P2×V2
Let's set the equation up by moving variables to solve what we are looking for, which is P2.)
P2=(P1×V1)/V2
Step 3: Plug in Values and Solve for P2: Using P2=(P1×V1)/V2
Given: P2=(P1×V1)/V2
P2=(44.7 PSIa × 4 ft³)/2 ft³
P2=89.4 PSIa
The new pressure (P2) of the gas is 89.4 PSIa after the volume is decreased to 2 cubic feet while the temperature remains constant.
So, in this example, using Boyle's Law, we determined that the pressure of the gas increases from 2 atm to 4 atm when the volume decreases from 4 ft³ to 2 ft³, assuming constant temperature.