The pressure of a certain gas in atmospheres, p, and its corresponding volume in liters, v, are shown in...

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The pressure of a certain gas in atmospheres, p, and its corresponding volume in liters, v, are shown in the table.
A two column table with 6 rows. The first row, v, has the entries, 1, 2, 3, 4, 5. The second column, p, has the entries, 3.00, 1.50, 1.00, 0.75, 0.60.
For the points in the table, does pressure vary directly with volume? If so, identify k and write an equation.
No, this is not a direct variation.
Yes, this is a direct variation with k = .
Yes, this is a direct variation with k = 1.
Yes, this is a direct variation with k = 3.

SnapAndSolve AI · Accepted Answer

To determine if the pressure $ p $ varies directly with the volume $ v $, we need to check if there is a constant $ k $ such that $ p = k \cdot v $ for the pairs of $ (v, p) $ from the table. The table entries are as follows: $$ \begin{array}{|c|c|} \hline v & p \ \hline 1 & 3.00 \ 2 & 1.50 \ 3 & 1.00 \ 4 & 0.75 \ 5 & 0.60 \ \hline \end{array} $$ Now we will calculate $ k $ for each pair $ (v, p) $: 1. For $ v = 1 $, $ p = 3.00 $: $$ k = \frac{p}{v} = \frac{3.00}{1} = 3.00 $$ 2. For $ v = 2 $, $ p = 1.50 $: $$ k = \frac{p}{v} = \frac{1.50}{2} = 0.75 $$ 3. For $ v = 3 $, $ p = 1.00 $: $$ k = \frac{p}{v} = \frac{1.00}{3} \approx 0.33 $$ 4. For $ v = 4 $, $ p = 0.75 $: $$ k = \frac{p}{v} = \frac{0.75}{4} = 0.1875 $$ 5. For $ v = 5 $, $ p = 0.60 $: $$ k = \frac{p}{v} = \frac{0.60}{5} = 0.12 $$ Since the value of $ k $ is not constant across the different pairs of $ (v, p) $, this indicates that pressure does not vary directly with volume. No, this is not a direct variation.

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The pressure of a certain gas in atmospheres, p, and its corresponding volume in liters, v, are shown in...

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