Question

A gas dissolves at 2.0 g/L at 1 atm. What pressure us needed to reach 3.0 g/L

Answer 1

To determine the pressure required to reach a higher concentration of a gas in solution, we can use Henry's Law, which states that the concentration (\(C\)) of a gas dissolved in a liquid is directly proportional to the pressure (\(P\)) of the gas above the liquid. This relationship can be expressed by the equation:

\[
C = k_H \cdot P
\]

Where:
- \(C\) is the concentration of the gas in g/L,
- \(k_H\) is Henry's law constant (in g/L·atm), and
- \(P\) is the pressure in atm.

From the information given, we know that at 1 atm, the concentration of the gas is 2.0 g/L.

We can find the Henry's law constant (\(k_H\)) using this data:

\[
k_H = \frac{C}{P} = \frac{2.0 \, \text{g/L}}{1 \, \text{atm}} = 2.0 \, \text{g/L·atm}
\]

Next, we want to find the pressure needed to achieve a concentration of 3.0 g/L. Using Henry's Law again, we can rearrange the formula to solve for pressure:

\[
P = \frac{C}{k_H}
\]

Substituting \(C = 3.0 \, \text{g/L}\) and \(k_H = 2.0 \, \text{g/L·atm}\):

\[
P = \frac{3.0 \, \text{g/L}}{2.0 \, \text{g/L·atm}} = 1.5 \, \text{atm}
\]

Therefore, the pressure needed to reach a concentration of 3.0 g/L is **1.5 atm**.