Question

A gas has a solubility of 1.46 g L at 8.00 atm of pressure. What is the pressure of a sample of the same gas that

contains 2.7 g/L of the dissolved gas?

Answer 1

To find the pressure of a sample containing 2.7 g/L of a dissolved gas using Henry's Law, you use the relationship between the initial solubility and pressure (1.46 g/L at 8.00 atm) to calculate the new pressure. The pressure is found to be approximately 14.79 atm.

Step-by-step explanation:

The student's question involves the application of Henry's Law, which describes the relationship between the solubility of a gas and the pressure of that gas above the solution. According to Henry's Law, solubility of a gas in a liquid at a given temperature is directly proportional to the pressure of that gas above the liquid. Specifically, the formula S1 / P1 = S2 / P2 can be used, where S represents solubility and P represents pressure.

In the given scenario, we have an initial solubility (S1) of 1.46 g/L at an initial pressure (P1) of 8.00 atm. We are then asked to find the pressure (P2) when the solubility (S2) is 2.7 g/L.

Using the formula from Henry's Law and solving for P2:

P2 = S2 × P1 / S1
P2 = 2.7 g/L × 8.00 atm / 1.46 g/L
P2 ≈ 14.79 atm

Therefore, the pressure of the sample when it contains 2.7 g/L of the dissolved gas is approximately 14.79 atm.

Answer 2

`14.8\ \text{atm}`

Step-by-step explanation:

`C_1` = Initial concentration = 1.46 g/L

`C_2` = Final concentration = 2.7 g/L

`P_1` = Initial pressure = 8 atm

`P_2` = Final pressure

From Henry's law we have the relation

`(C_2)/(C_1)=(P_2)/(P_1)\\\Rightarrow P_2=(C_2)/(C_1)P_1\\\Rightarrow P_2=(2.7)/(1.46)* 8\\\Rightarrow P_2=14.8\ \text{atm}`

The pressure of a sample of the same gas at the required concentration is
`14.8\ \text{atm}`.