Question

use the equation $p = \frac{8.31}{v}$, where $p = $ pressure and $v = $ volume. what happens to the pressure as the volume approaches 0? the pressure approaches one. the pressure approaches negative infinity. the pressure approaches zero. the pressure approaches infinity.

Explanation:

Step1: Analyze the function type

The equation \( p=\frac{8.31}{V} \) is a rational function, where \( p \) is pressure and \( V \) is volume. Since volume \( V \) in a physical context (like gas volume) is positive ( \( V>0 \) ), we consider the limit as \( V \) approaches 0 from the positive side ( \( V
ightarrow0^{+} \) ).

Step2: Evaluate the limit

For the function \( y = \frac{k}{x} \) where \( k>0 \) and \( x>0 \), as \( x \) approaches 0 from the right ( \( x
ightarrow0^{+} \) ), the value of \( y \) increases without bound (approaches positive infinity). In our case, \( k = 8.31>0 \) and \( V>0 \) (volume can't be negative in this context). So as \( V
ightarrow0^{+} \), \( p=\frac{8.31}{V}\) will approach positive infinity (since dividing a positive constant by a positive number that gets closer and closer to 0 makes the result get larger and larger).

Answer:

The pressure approaches infinity.