High School

Gas Laws Fact Sheet

[tex]
\[
\begin{tabular}{|c|c|}
\hline
Ideal gas law & $PV = nRT$ \\
\hline
Ideal gas constant &
\begin{tabular}{l}
$R = 8.314 \frac{L \cdot kPa}{mol \cdot K}$ \\
or \\
$R = 0.0821 \frac{L \cdot atm}{mol \cdot K}$
\end{tabular} \\
\hline
Standard atmospheric pressure & $1 \text{ atm} = 101.3 \text{ kPa}$ \\
\hline
\end{tabular}
\]
[/tex]

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. 1.5 kPa
B. 241 kPa
C. 25.6 kPa
D. 112.6 kPa
E. 225.2 kPa

Answer :

To find the gauge pressure, we use the definition that gauge pressure is the difference between the absolute pressure inside the container and the ambient (atmospheric) pressure outside. This can be expressed as:

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

Given:
- Absolute pressure in the container: [tex]$P_{\text{absolute}} = 125.4\, \text{kPa}$[/tex]
- Ambient pressure around the container: [tex]$P_{\text{ambient}} = 99.8\, \text{kPa}$[/tex]

Substitute these values into the formula:

[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]$\boxed{25.6\, \text{kPa}}$[/tex], which corresponds to option C.

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