MathAdmin: Created page with "Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure $P$ an..."

2017-04-10T16:31:57Z

Created page with "<span class="exam">Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure <math style="vertical-align: 0px">P</math> an..."

New page

<span class="exam">Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure  <math style="vertical-align: 0px">P</math>  and volume  <math style="vertical-align: 0px">V</math>  satisfy the equation  <math style="vertical-align: 0px">PV=C</math>  where  <math style="vertical-align: 0px">C</math>  is a constant. Suppose that at a certain instant, the volume is  <math style="vertical-align: -4px">600 \text{ cm}^3,</math>  the pressure is  <math style="vertical-align: -4px">150 \text{ kPa},</math>  and the pressure is increasing at a rate of  <math style="vertical-align: -4px">20 \text{ kPa/min}.</math>  At what rate is the volume decreasing at this instant?

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''Product Rule'''
|-
|        <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we take the derivative of the equation  <math style="vertical-align: 0px">PV=C.</math>
|-
|Using the product rule, we get
|-
|       <math>P'V+PV'=C'.</math>
|-
|Since  <math style="vertical-align: 0px">C</math>  is a constant,
|-
|       <math style="vertical-align: -1px">C'=0.</math> 
|-
|Therefore, we have
|-
|       <math>P'V+PV'=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Solving for  <math style="vertical-align: -4px">V',</math>  we get
|-
|       <math>V'=\frac{-P'V}{P}.</math>
|-
|Using the information provided in the problem, we have
|-
|       <math style="vertical-align: 0px">V=600 \text{ cm}^3,~P=150 \text{ kPa},~P'=20 \text{ kPa/min}.</math> 
|-
|Hence, we get
|-
|
        <math>\begin{array}{rcl}
\displaystyle{V'} & = & \displaystyle{\frac{-(20)(600)}{150} \text{ cm}^3\text{/min}}\\
&&\\
& = & \displaystyle{-80 \text{ cm}^3\text{/min}.}
\end{array}</math>
|-
|Therefore, the volume is decreasing at a rate of  <math style="vertical-align: -5px">80 \text{ cm}^3\text{/min}</math>  at this instant.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       The volume is decreasing at a rate of  <math style="vertical-align: -5px">80 \text{ cm}^3\text{/min}</math>  at this instant.
|}
[[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

009A Sample Final 3, Problem 8 - Revision history

MathAdmin: Created page with "Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P an..."

MathAdmin: Created page with "Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure $P$ an..."