Q6) A 10 L container contains a mixture of 0.8 g of He and 8 g of Ar at 0 oC.
a) (1 pt) Calculate the total pressure of the gas mixture.
b) (1 pt) Calculate the partial pressure of each gas in the gas mixture.
c) (1 pt) If the initial total pressure of the container was 1.5 atm (at 0 oC), what will be the new pressure if the container is heated to 100 oC (Volume is kept constant)?
d) (1 pt) If the initial total pressure of the container was 1.5 atm (with 0.8 g He + 8 g Ar), what will be the new pressure in the container if 1 mol of Ne is added to this gas mixture
(Volume is kept constant).
a. Convert g He and g Ar to moles. mols = g/molar mass. Add moles and use total moles and PV = nRT to solve for total P.
b. Use PV = nRT and individual n values for partial pressure of each gas.
c.(P1/T1) = (P2/T2)
d. sum n He + n Ar + 1 mol for Ne. Then PV = nRT
a) To calculate the total pressure of the gas mixture, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
In this case, we are given the volume (10 L), and we need to calculate the total pressure of the gas mixture. Let's calculate the number of moles for each gas first.
For He:
Given mass = 0.8 g
Molar mass of He = 4 g/mol
n(He) = mass/Molar mass = 0.8 g/4 g/mol = 0.2 mol
For Ar:
Given mass = 8 g
Molar mass of Ar = 40 g/mol
n(Ar) = mass/Molar mass = 8 g/40 g/mol = 0.2 mol
Total number of moles (n) = n(He) + n(Ar) = 0.2 mol + 0.2 mol = 0.4 mol
Now, we can substitute the values into the ideal gas law equation to find the total pressure:
P(total) * V = n * R * T
P(total) = (n * R * T) / V
P(total) = (0.4 mol * R * T) / 10 L
R is the ideal gas constant, which is 0.0821 L atm/(mol K).
T is the temperature in Kelvin. Given 0°C, we need to convert it to Kelvin: 0°C + 273.15 = 273.15 K.
P(total) = (0.4 mol * 0.0821 L atm/(mol K) * 273.15 K) / 10 L
P(total) ≈ 8.898 atm
Therefore, the total pressure of the gas mixture is approximately 8.898 atm.
b) To calculate the partial pressure of each gas in the mixture, we can use the following formula:
Partial pressure = (number of moles of the gas / total number of moles) * total pressure
For He:
Partial pressure(He) = (n(He) / n) * P(total)
Partial pressure(He) = (0.2 mol / 0.4 mol) * 8.898 atm
Partial pressure(He) ≈ 4.449 atm
For Ar:
Partial pressure(Ar) = (n(Ar) / n) * P(total)
Partial pressure(Ar) = (0.2 mol / 0.4 mol) * 8.898 atm
Partial pressure(Ar) ≈ 4.449 atm
Therefore, the partial pressure of both He and Ar in the mixture is approximately 4.449 atm.
c) To find the new pressure when the temperature is changed to 100°C while keeping the volume constant, we can use the relationship between temperature and pressure known as Gay-Lussac's law:
P1 / T1 = P2 / T2
Where:
P1 = initial pressure
T1 = initial temperature
P2 = final pressure
T2 = final temperature
Given:
P1 = 1.5 atm
T1 = 0°C + 273.15 = 273.15 K
T2 = 100°C + 273.15 = 373.15 K
Substituting the values into the equation, we can solve for P2:
P2 = (P1 * T2) / T1
P2 = (1.5 atm * 373.15 K) / 273.15 K
P2 ≈ 2.05 atm
Therefore, the new pressure when the container is heated to 100°C (volume kept constant) is approximately 2.05 atm.
d) To calculate the new pressure when 1 mol of Ne is added to the gas mixture while keeping the volume constant, we can use Dalton's law of partial pressures:
P(total new) = P(total initial) + P(Ne)
Given:
P(total initial) = 1.5 atm
P(Ne) = ?
Since Ne is added, the number of moles of Ne is 1 mol. We already calculated the total number of moles in the initial mixture, which is 0.4 mol. Therefore, we have:
P(total new) = 1.5 atm + P(Ne)
We need to solve for P(Ne).
P(total new) = P(total initial) + (n(Ne) / n) * P(total initial)
P(total new) = 1.5 atm + (1 mol / 0.4 mol) * 1.5 atm
P(total new) ≈ 5.25 atm
Therefore, the new pressure in the container, when 1 mol of Ne is added to the gas mixture while keeping the volume constant, is approximately 5.25 atm.