Math / Algebra

QuestionThe volume of a gas varies directly with its temperature and inversely with its pressure. The volume of a gas is $768 \mathrm{~m}^{3}$ when the temperature is 268 K and pressure is 29 Pa . What is the volume, in cubic meters, of the gas when the temperature is 325 K and pressure is 11 Pa ? Round to the nearest integer.

Studdy Solution

STEP 1

What is this asking? If a gas's volume changes with temperature and pressure in a specific way, and we know its volume at one point, what's its volume at another point with a different temperature and pressure? Watch out! Don't mix up direct and inverse variation – they work in opposite ways!
Also, remember to round to the nearest integer at the end.

STEP 2

1. Set up the combined variation equation.
2. Find the constant of variation.
3. Calculate the new volume.

STEP 3

Alright, so we know the volume ( $V$ ) varies directly with temperature ( $T$ ) and inversely with pressure ( $P$ ).
Direct variation means they change in the same direction – if one goes up, the other goes up!
Inverse variation means they change in opposite directions.

STEP 4

We can write this relationship as $V = k \cdot \frac{T}{P}$ , where $k$ is our constant of variation.
This equation captures both the direct relationship with $T$ and the inverse relationship with $P$ .

STEP 5

We're given an initial state: $V = \textbf{768} \mathrm{~m}^{3}$ , $T = \textbf{268}$ K, and $P = \textbf{29}$ Pa.
Let's plug these values into our equation: $\textbf{768} = k \cdot \frac{\textbf{268}}{\textbf{29}}$ .

STEP 6

To solve for $k$ , we can multiply both sides by $\frac{\textbf{29}}{\textbf{268}}$ .
This gives us $k = \textbf{768} \cdot \frac{\textbf{29}}{\textbf{268}}$ .

STEP 7

Calculating this gives us $k \approx \textbf{83.02985}$ .
We'll keep a few extra decimal places for accuracy and round at the very end.

STEP 8

Now we know $k$ , and we're given a new temperature and pressure: $T = \textbf{325}$ K and $P = \textbf{11}$ Pa.
We can plug these, along with our calculated $k$ , back into our equation: $V = \textbf{83.02985} \cdot \frac{\textbf{325}}{\textbf{11}}$ .

STEP 9

Calculating this gives us $V \approx \textbf{2460.2985}$ $\mathrm{~m}^{3}$ .

STEP 10

Finally, we need to round to the nearest integer, which gives us a final volume of approximately $\textbf{2460}$ $\mathrm{~m}^{3}$ .

STEP 11

The volume of the gas at 325 K and 11 Pa is approximately 2460 $\mathrm{~m}^{3}$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the combined variation equation.
2. Find the constant of variation.
3. Calculate the new volume.

STEP 3

Alright, so we know the volume ( $V$ ) varies directly with temperature ( $T$ ) and inversely with pressure ( $P$ ).
Direct variation means they change in the same direction – if one goes up, the other goes up!
Inverse variation means they change in opposite directions.

STEP 4

We can write this relationship as $V = k \cdot \frac{T}{P}$ , where $k$ is our constant of variation.
This equation captures both the direct relationship with $T$ and the inverse relationship with $P$ .

STEP 5

We're given an initial state: $V = \textbf{768} \mathrm{~m}^{3}$ , $T = \textbf{268}$ K, and $P = \textbf{29}$ Pa.
Let's plug these values into our equation: $\textbf{768} = k \cdot \frac{\textbf{268}}{\textbf{29}}$ .

STEP 6

To solve for $k$ , we can multiply both sides by $\frac{\textbf{29}}{\textbf{268}}$ .
This gives us $k = \textbf{768} \cdot \frac{\textbf{29}}{\textbf{268}}$ .

STEP 7

Calculating this gives us $k \approx \textbf{83.02985}$ .
We'll keep a few extra decimal places for accuracy and round at the very end.

STEP 8

Now we know $k$ , and we're given a new temperature and pressure: $T = \textbf{325}$ K and $P = \textbf{11}$ Pa.
We can plug these, along with our calculated $k$ , back into our equation: $V = \textbf{83.02985} \cdot \frac{\textbf{325}}{\textbf{11}}$ .

STEP 9

Calculating this gives us $V \approx \textbf{2460.2985}$ $\mathrm{~m}^{3}$ .

STEP 10

Finally, we need to round to the nearest integer, which gives us a final volume of approximately $\textbf{2460}$ $\mathrm{~m}^{3}$ .

STEP 11

The volume of the gas at 325 K and 11 Pa is approximately 2460 $\mathrm{~m}^{3}$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the combined variation equation.2. Find the constant of variation.3. Calculate the new volume.

STEP 3

Alright, so we know the volume (VVV) varies *directly* with temperature (TTT) and *inversely* with pressure (PPP).Direct variation means they change in the same direction – if one goes up, the other goes up!Inverse variation means they change in opposite directions.

STEP 4

We can write this relationship as V=k⋅TPV = k \cdot \frac{T}{P}V=k⋅PT​, where kkk is our **constant of variation**.This equation captures both the direct relationship with TTT and the inverse relationship with PPP.

STEP 5

We're given an **initial state**: V=768 m3V = \textbf{768} \mathrm{~m}^{3}V=768 m3, T=268T = \textbf{268}T=268 K, and P=29P = \textbf{29}P=29 Pa.Let's plug these values into our equation: 768=k⋅26829\textbf{768} = k \cdot \frac{\textbf{268}}{\textbf{29}}768=k⋅29268​.

STEP 6

To solve for kkk, we can multiply both sides by 29268\frac{\textbf{29}}{\textbf{268}}26829​.This gives us k=768⋅29268k = \textbf{768} \cdot \frac{\textbf{29}}{\textbf{268}}k=768⋅26829​.

STEP 7

Calculating this gives us k≈83.02985k \approx \textbf{83.02985}k≈83.02985.We'll keep a few extra decimal places for accuracy and round at the very end.

STEP 8

Now we know kkk, and we're given a new temperature and pressure: T=325T = \textbf{325}T=325 K and P=11P = \textbf{11}P=11 Pa.We can plug these, along with our calculated kkk, back into our equation: V=83.02985⋅32511V = \textbf{83.02985} \cdot \frac{\textbf{325}}{\textbf{11}}V=83.02985⋅11325​.

STEP 9

Calculating this gives us V≈2460.2985V \approx \textbf{2460.2985}V≈2460.2985 m3\mathrm{~m}^{3} m3.

STEP 10

Finally, we need to round to the nearest integer, which gives us a **final volume** of approximately 2460\textbf{2460}2460 m3\mathrm{~m}^{3} m3.

STEP 11

The volume of the gas at 325 K and 11 Pa is approximately 2460 m3\mathrm{~m}^{3} m3.

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1. Set up the combined variation equation.
2. Find the constant of variation.
3. Calculate the new volume.

Alright, so we know the volume ( $V$ ) varies directly with temperature ( $T$ ) and inversely with pressure ( $P$ ).
Direct variation means they change in the same direction – if one goes up, the other goes up!
Inverse variation means they change in opposite directions.

We can write this relationship as $V = k \cdot \frac{T}{P}$ , where $k$ is our constant of variation.
This equation captures both the direct relationship with $T$ and the inverse relationship with $P$ .

We're given an initial state: $V = \textbf{768} \mathrm{~m}^{3}$ , $T = \textbf{268}$ K, and $P = \textbf{29}$ Pa.
Let's plug these values into our equation: $\textbf{768} = k \cdot \frac{\textbf{268}}{\textbf{29}}$ .

To solve for $k$ , we can multiply both sides by $\frac{\textbf{29}}{\textbf{268}}$ .
This gives us $k = \textbf{768} \cdot \frac{\textbf{29}}{\textbf{268}}$ .

Calculating this gives us $k \approx \textbf{83.02985}$ .
We'll keep a few extra decimal places for accuracy and round at the very end.

Now we know $k$ , and we're given a new temperature and pressure: $T = \textbf{325}$ K and $P = \textbf{11}$ Pa.
We can plug these, along with our calculated $k$ , back into our equation: $V = \textbf{83.02985} \cdot \frac{\textbf{325}}{\textbf{11}}$ .

Calculating this gives us $V \approx \textbf{2460.2985}$ $\mathrm{~m}^{3}$ .

Finally, we need to round to the nearest integer, which gives us a final volume of approximately $\textbf{2460}$ $\mathrm{~m}^{3}$ .

The volume of the gas at 325 K and 11 Pa is approximately 2460 $\mathrm{~m}^{3}$ .