High School

Gas Laws Fact Sheet

[tex]
\[
\begin{array}{|l|l|}
\hline
\text{Ideal gas law} & P V = n R T \\
\hline
\text{Ideal gas constant} &
\begin{array}{l}
R = 8.314 \frac{L \cdot kPa}{mol \cdot K} \\
\text{or} \\
R = 0.0821 \frac{L \cdot atm}{mol \cdot K}
\end{array} \\
\hline
\text{Standard atmospheric pressure} & 1 \, atm = 101.3 \, kPa \\
\hline
\text{Celsius to Kelvin conversion} & K = {}^{\circ} C + 273.15 \\
\hline
\end{array}
\]
[/tex]

Select the correct answer.

The gas in a sealed container has an absolute pressure of 125.4 kilopascals. If the air around the container is at a pressure of 99.8 kilopascals, what is the gauge pressure inside the container?

A. [tex]\(1.5 \, kPa\)[/tex]

B. [tex]\(24.1 \, kPa\)[/tex]

C. [tex]\(25.6 \, kPa\)[/tex]

D. [tex]\(112.6 \, kPa\)[/tex]

E. [tex]\(225.2 \, kPa\)[/tex]

Answer :

To find the gauge pressure inside the container, we use the fact that the gauge pressure is the difference between the absolute pressure inside the container and the atmospheric pressure outside. That is, the formula is:

[tex]$$
P_{\text{gauge}} = P_{\text{absolute}} - P_{\text{atmospheric}}
$$[/tex]

Given that the absolute pressure is [tex]$125.4\ \text{kPa}$[/tex] and the atmospheric pressure is [tex]$99.8\ \text{kPa}$[/tex], we substitute these values into the equation:

[tex]$$
P_{\text{gauge}} = 125.4\ \text{kPa} - 99.8\ \text{kPa}
$$[/tex]

Performing the subtraction:

[tex]$$
P_{\text{gauge}} = 25.6\ \text{kPa}
$$[/tex]

Thus, the gauge pressure inside the container is [tex]$25.6\ \text{kPa}$[/tex]. The correct answer is Option C.

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