When a car tire is inflated to a volume of 10 L of air, it has a pressure of 2.2 atm. If the volume is inversely proportional to the pressure, then what is the volume of air needed to inflate a tire to a pressure of 2.1 atm? Round to one decimal.

Given that the volume \(V\) is inversely proportional to the pressure \(P\), we can express this relationship as:

\[ V \propto \frac{1}{P} \]

This can be written as:

\[ V \cdot P = k \]

where \(k\) is a constant.

First, we need to find the constant \(k\) using the initial conditions:
\[ V_1 = 10 \text{ L} \]
\[ P_1 = 2.2 \text{ atm} \]

Plugging these values into the equation, we get:

\[ k = V_1 \cdot P_1 = 10 \text{ L} \cdot 2.2 \text{ atm} = 22 \text{ L} \cdot \text{atm} \]

Next, we need to find the volume \(V_2\) when the pressure \(P_2\) is 2.1 atm. Using the constant \(k\), we have:

\[ V_2 \cdot P_2 = k \]

\[ V_2 \cdot 2.1 \text{ atm} = 22 \text{ L} \cdot \text{atm} \]

Solving for \(V_2\):

\[ V_2 = \frac{22 \text{ L} \cdot \text{atm}}{2.1 \text{ atm}} \]

\[ V_2 \approx 10.5 \text{ L} \]

So, the volume of air needed to inflate the tire to a pressure of 2.1 atm is approximately 10.5 liters.

Would you like more details or have any questions?

Here are 5 questions you might want to ask:
1. How do you derive the relationship \( V \propto \frac{1}{P} \)?
2. Can you explain the concept of inverse proportionality in more detail?
3. What are the units used in this type of calculation?
4. How does temperature affect the pressure-volume relationship?
5. Can this principle be applied to other gases besides air?

Tip: When solving problems involving proportionality, always remember to identify the constant of proportionality first, as it simplifies finding the unknowns.

Learn how to calculate the volume of air needed to inflate a car tire to a pressure of 2.1 atm from given conditions of inverse proportionality between volume and pressure. Detailed step-by-step solution provided.

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