Use Henry's law to determine the molar solubility of helium at a pressure of 1.5 atm and 25deg C.
To determine the molar solubility of helium at a given pressure and temperature using Henry's law, we need to know the Henry's law constant, which represents the solubility of a gas in a solvent at a particular temperature.
Henry's law states that the solubility of a gas is directly proportional to the partial pressure of the gas above the solvent:
C = k * P
Where:
C is the molar concentration (molarity) of the gas in the solution,
k is the Henry's law constant, and
P is the partial pressure of the gas.
In this case, we need to find k, the Henry's law constant for helium at 25°C. The Henry's law constant varies depending on the solvent and temperature, so a specific value needs to be given or looked up.
Let's assume the given Henry's law constant for helium in water at 25°C is 4.5 x 10^(-4) mol/(L·atm).
Now, we can substitute the values into Henry's law equation to find the molar solubility of helium (C):
C = (k * P) / R * T
Where:
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
T is the temperature in Kelvin (25°C = 298 K). Please note that temperatures should be in Kelvin when working with gas laws.
C = (4.5 x 10^(-4) mol/(L·atm) * 1.5 atm) / (0.0821 L·atm/(mol·K) * 298 K)
C ≈ 9.1 x 10^(-5) mol/L
Therefore, the molar solubility of helium at a pressure of 1.5 atm and 25°C is approximately 9.1 x 10^(-5) mol/L.