At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Part A)
According to the text, the equation that relates x and y is:
[tex]y=\frac{14}{x}[/tex]Substitute x=21 to find the value of y when x=21:
[tex]y=\frac{14}{21}[/tex]Simplify the expression:
[tex]\frac{14}{21}=\frac{7\cdot2}{7\cdot3}=\frac{2}{3}[/tex]Therefore, the value of y when x=21 is:
[tex]\frac{2}{3}[/tex]Part B)
To find the value of x when y=28, substitute y=28 and solve for x:
[tex]\begin{gathered} y=\frac{14}{x} \\ \Rightarrow28=\frac{14}{x} \\ \Rightarrow28x=14 \\ \Rightarrow x=\frac{14}{28} \\ \therefore x=\frac{1}{2} \end{gathered}[/tex]Therefore, the value of x when y=28 is:
[tex]\frac{1}{2}[/tex]Part C)
Since the pressure P is inversely proportional to the volume V, then:
[tex]P=\frac{k}{V}[/tex]Solve for k and substitute P=250kPa and V=1.7m^2 to find the constant of proportionality:
[tex]\begin{gathered} \Rightarrow k=PV \\ =(250\text{kPa})(1.7m^3) \\ =425\text{kPa}\cdot m^3 \end{gathered}[/tex]Therefore, the constant of proportionality for this situation is:
[tex]425\text{ kPa}\cdot m^3[/tex]Part D)
Substitute the value of k into the equation that shows the inverse relation between P and V:
[tex]\begin{gathered} P=\frac{k}{V} \\ \Rightarrow P=\frac{425\text{ kPa}\cdot m^3}{V} \end{gathered}[/tex]Therefore, the inverse variation equation model for this situation, is:
[tex]P=\frac{425\text{ kPa}\cdot m^3}{V}[/tex]Part E)
Substitute V=3.2m^3 to find the pressure under those conditions:
[tex]\begin{gathered} P=\frac{425\text{ kPa}\cdot m^3}{3.2m^3} \\ =132.8\text{kPa} \end{gathered}[/tex]Therefore, the pressure would be:
[tex]132.8\text{kPa}[/tex]Part F)
Isolate V from the equation and substitute P=150kPa:
[tex]\begin{gathered} V=\frac{425\text{ kPa}\cdot m^3}{P} \\ =\frac{425\text{ kPa}\cdot m^3}{150\text{ kPa}} \\ =2.83m^3 \end{gathered}[/tex]Therefore, the approximate volume would have to be equal to:
[tex]2.83m^3[/tex]Part G)
Since the temperature has decreased, the pressure must be lower according to the description provided in the text. Then, an inequality to model this situation would be:
[tex]P<\frac{k}{V}[/tex]Part H)
Substitute the value of k and V=3m^3:
[tex]\begin{gathered} P<\frac{425\text{ kPa}\cdot m^3}{3m^3} \\ \Rightarrow P<141.7\text{ kPa}^{} \end{gathered}[/tex]Part I)
Mathematically, all numbers under 141.7 satisfy the inequality from part H. Nevertheless, negative pressures do not have a physical meaning under the context of the Ideal Gas Law. Therefore, we must include the condition that P is greater than 0:
[tex]0
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.