Gas Laws Fact Sheet

[tex]\[

\begin{array}{|l|l|}

\hline

\text{Ideal gas law} & PV = nRT \\

\hline

\text{Ideal gas constant} &

\begin{array}{l}

R = 8.314 \frac{L \cdot \text{kPa}}{\text{mol} \cdot K} \\

\text{or} \\

R = 0.0821 \frac{L \cdot \text{atm}}{\text{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}

\][/tex]

Select the correct answer.

When a chemist collects hydrogen gas over water, she ends up with a mixture of hydrogen and water vapor in her collecting bottle. If the pressure in the collecting bottle is 97.1 kilopascals and the vapor pressure of the water is 3.2 kilopascals, what is the partial pressure of the hydrogen?

A. 93.9 kPa
B. 98.1 kPa
C. 100.3 kPa
D. 104.5 kPa

When gases are collected over water, the total pressure in the container is the sum of the partial pressures of the gas of interest and the water vapor. Mathematically, this is expressed as

[tex]$$
P_{\text{total}} = P_{\text{gas}} + P_{\text{H}_2O}.
$$[/tex]

To find the partial pressure of the hydrogen gas, we rearrange the equation:

[tex]$$
P_{\text{gas}} = P_{\text{total}} - P_{\text{H}_2O}.
$$[/tex]

Given:
- The total pressure [tex]$P_{\text{total}} = 97.1 \, \text{kPa}$[/tex],
- The vapor pressure of water [tex]$P_{\text{H}_2O} = 3.2 \, \text{kPa}$[/tex],

we substitute these values into the equation:

[tex]$$
P_{\text{gas}} = 97.1 \, \text{kPa} - 3.2 \, \text{kPa} = 93.9 \, \text{kPa}.
$$[/tex]

Thus, the partial pressure of the hydrogen gas is

[tex]$$
\boxed{93.9 \, \text{kPa}},
$$[/tex]

which corresponds to option A.