High School

Gas Laws Fact Sheet

\[
\begin{array}{|l|l|}
\hline
\text{Ideal gas law} & PV = nRT \\
\hline
\text{Ideal gas constant} & \begin{array}{l}
R = 8.314 \text{ (J/mol·K)} \\
\text{or} \\
R = 0.0821 \frac{L \cdot atm}{mol \cdot K}
\end{array} \\
\hline
\text{Standard atmospheric pressure} & 1 \text{ atm} = 101.3 \text{ kPa} \\
\hline
\text{Celsius to Kelvin conversion} & K = {^\circ}C + 273.15 \\
\hline
\end{array}
\]

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. 112.0 kPa
E. 225.2 kPa

Answer :

The problem asks for the gauge pressure, which is the difference between the absolute pressure inside the container and the atmospheric pressure outside.

Step 1: Identify the pressures given.
- The absolute pressure inside the container is
$$P_{\text{abs}} = 125.4 \text{ kPa}.$$
- The atmospheric pressure is
$$P_{\text{atm}} = 99.8 \text{ kPa}.$$

Step 2: Write down the formula for gauge pressure.
The gauge pressure is calculated by subtracting the atmospheric pressure from the absolute pressure:

$$
P_{\text{gauge}} = P_{\text{abs}} - P_{\text{atm}}.
$$

Step 3: Substitute the given values into the formula:

$$
P_{\text{gauge}} = 125.4 \text{ kPa} - 99.8 \text{ kPa}.
$$

Step 4: Perform the subtraction:

$$
P_{\text{gauge}} = 25.6 \text{ kPa}.
$$

Thus, the gauge pressure inside the container is $$25.6 \text{ kPa}.$$

This corresponds to option C.

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