Question

use the equation (p = \frac{8.31}{v}) where (p = \text{pressure}) and (v = \text{volume}).

what happens to the pressure as the volume approaches 0? explain your reasoning.

Explanation:

🆕 New Concept Discovered: Inverse Variation and Limits
As the denominator gets closer to zero, the fraction grows without bound.

Step 1: Analyze the relationship

The given equation is:

$$ p = \frac{8.31}{V} $$

In this equation, pressure \( p \) is inversely proportional to volume \( V \). This means that as one variable increases, the other decreases, and vice versa.

Step 2: Evaluate the behavior as \( V \) approaches 0

To find what happens to the pressure \( p \) as the volume \( V \) approaches \( 0 \) from the positive side (since volume must be positive, \( V \to 0^+ \)), we look at the behavior of the fraction:

• If \( V = 1 \), then \( p = \frac{8.31}{1} = 8.31 \)
• If \( V = 0.1 \), then \( p = \frac{8.31}{0.1} = 83.1 \)
• If \( V = 0.01 \), then \( p = \frac{8.31}{0.01} = 831 \)
• If \( V = 0.001 \), then \( p = \frac{8.31}{0.001} = 8310 \)

As the denominator \( V \) becomes an extremely small positive number close to \( 0 \), the value of the fraction \( \frac{8.31}{V} \) grows larger and larger without bound.

Step 3: Formulate the conclusion

Mathematically, this limit is expressed as:

$$ \lim_{V \to 0^+} \frac{8.31}{V} = \infty $$

Therefore, as the volume approaches \( 0 \), the pressure increases toward infinity (grows infinitely large).

Answer:

As the volume \( V \) approaches \( 0 \), the pressure \( p \) increases toward infinity (grows infinitely large).

Reasoning:
In the equation \( p = \frac{8.31}{V} \), pressure and volume are inversely proportional. Dividing a constant positive number (\( 8.31 \)) by an increasingly smaller positive value that gets closer and closer to \( 0 \) results in an increasingly larger quotient. Therefore, as \( V \to 0^+ \), the pressure \( p \) increases without bound toward infinity (\( \infty \)).