College

**Gas Laws Fact Sheet**

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

**Question:**

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. 1.5 kPa
B. 24.1 kPa
C. 25.6 kPa
D. 112.6 kPa
E. 225.2 kPa

Answer :

To determine the gauge pressure inside the container, we use the definition of gauge pressure, which is the difference between the absolute pressure inside the container and the atmospheric pressure outside. Mathematically, this is given by:

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

Given the values:

- Absolute pressure: [tex]$$P_{\text{absolute}} = 125.4 \, \text{kPa}$$[/tex]
- Atmospheric pressure: [tex]$$P_{\text{atmospheric}} = 99.8 \, \text{kPa}$$[/tex]

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], which corresponds to option C.

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