If the volume is decreased by 20% at a constant temperature, what is the change
in the pressure?

According to **Boyle's Law**, for a given mass of gas at constant temperature, the pressure of a gas is inversely proportional to its volume. Mathematically, this relationship is expressed as:

\[
P_1 V_1 = P_2 V_2
\]

Where:
- \(P_1\) is the initial pressure,
- \(V_1\) is the initial volume,
- \(P_2\) is the final pressure,
- \(V_2\) is the final volume.

Given that the volume is decreased by 20%, the new volume \(V_2\) becomes:

\[
V_2 = 0.8 V_1
\]

Substitute this into Boyle's Law equation:

\[
P_1 V_1 = P_2 (0.8 V_1)
\]

Simplifying, we get:

\[
P_2 = \frac{P_1 V_1}{0.8 V_1} = \frac{P_1}{0.8}
\]

Therefore:

\[
P_2 = 1.25 P_1
\]

This means the pressure increases by 25%.

### Summary:
If the volume is decreased by 20% at constant temperature, the pressure increases by 25%.

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Do you need further details or have any questions? Here are some related problems to explore:

1. How would the pressure change if the volume is reduced by 50%?
2. What happens to the pressure if the volume increases by 10% at constant temperature?
3. How does Boyle's Law apply to real gases compared to ideal gases?
4. If the temperature is also reduced, how does that affect Boyle's Law?
5. Can you derive Boyle's Law from the ideal gas law?

**Tip:** When working with gas laws, always remember to keep units consistent and ensure you're working with absolute pressure and volume values.

If the volume is decreased by 20% at a constant temperature, what is the change in the pressure?

Understand how to calculate the change in pressure when the volume decreases by 20% at constant temperature using Boyle's Law. Learn the key formulas, including P1 * V1 = P2 * V2 and how the pressure increases by 25%. Suitable for high school students in Grades 9-11.

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