a sample containing 1.50 mol neon gas has a volume of 8.00L.what is the new volume of the gas in liters when the following changes occur in the quantity of the gas at constant pressure and temperature?
a.a leak allows on-half of the neon atoms to escape.
b. a sample of 25.0g neon is added to the neon gas already in the container.
c a sample of 3.50 mol Ne is added to the neon gas already in the container.
Use PV = nRT
1.07 moles
To solve this problem, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (constant)
V = volume
n = number of moles
R = ideal gas constant
T = temperature (constant)
Given:
Number of moles, n = 1.50 mol
Initial volume, V = 8.00 L
a. A leak allows one-half of the neon atoms to escape:
In this case, the number of moles is reduced to half, which means n = 1.50 mol / 2 = 0.75 mol.
Since the pressure and temperature are constant, we can use the equation:
V1/n1 = V2/n2
Where:
V1 = initial volume
n1 = initial number of moles
V2 = new volume
n2 = new number of moles
Plugging in the values, we have:
V1/n1 = V2/n2
8.00 L / 1.50 mol = V2 / 0.75 mol
Simplifying the equation:
(8.00 L)(0.75 mol) = (1.50 mol)(V2)
6.00 L = 1.50 mol V2
Solving for V2:
V2 = 6.00 L / 1.50 mol
V2 = 4.00 L
Therefore, the new volume of the gas is 4.00 L when half of the neon atoms escape.
b. A sample of 25.0g neon is added to the neon gas already in the container:
To find the new volume, we need to calculate the number of moles for the added mass of neon.
Given:
Mass, m = 25.0 g (neon)
Molar mass, M = 20.18 g/mol (neon)
Number of moles, n2 = mass / molar mass
n2 = 25.0 g / 20.18 g/mol
n2 = 1.24 mol
Since the pressure and temperature are constant, we can use the equation:
V1/n1 = V2/n2
Plugging in the values:
8.00 L / 1.50 mol = V2 / (1.50 mol + 1.24 mol)
8.00 L / 1.50 mol = V2 / 2.74 mol
Simplifying and solving for V2:
(8.00 L)(2.74 mol) = (1.50 mol + 1.24 mol)(V2)
21.92 L = 2.74 mol V2
V2 = 21.92 L / 2.74 mol
V2 = 8.00 L
Therefore, the new volume of the gas remains 8.00 L when 25.0 g neon is added to the container.
c. A sample of 3.50 mol Ne is added to the neon gas already in the container:
To find the new volume, we need to add the number of moles from the new sample to the initial moles.
Given:
Number of moles, n2 = 3.50 mol
Again, using the equation:
V1/n1 = V2/n2
Plugging in the values:
8.00 L / 1.50 mol = V2 / (1.50 mol + 3.50 mol)
8.00 L / 1.50 mol = V2 / 5.00 mol
Simplifying and solving for V2:
(8.00 L)(5.00 mol) = (1.50 mol + 3.50 mol)(V2)
40.00 L = 5.00 mol V2
V2 = 40.00 L / 5.00 mol
V2 = 8.00 L
Therefore, the new volume of the gas remains 8.00 L when 3.50 mol neon is added to the container.
To find the new volume of the gas in each scenario, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature in Kelvin
Given:
Initial volume (V₁) = 8.00 L
Initial number of moles (n₁) = 1.50 mol
a. When half of the neon atoms escape, the new number of moles (n₂) will be half of the initial number of moles (n₁/2).
Calculating the new volume (V₂):
V₁ = (n₁ * R * T) / P
V₂ = (n₂ * R * T) / P
V₂ = [(n₁/2) * R * T] / P
Since the pressure (P) and temperature (T) remain constant, we can simplify the equation:
V₂ = (V₁ * n₁) / (2 * n₁)
V₂ = V₁ / 2
V₂ = 8.00 L / 2
V₂ = 4.00 L
Therefore, the new volume of the gas after half of the neon atoms escape is 4.00 L.
b. When 25.0 g of neon is added, we need to convert the mass of neon to moles using the molar mass of neon.
Molar mass of neon (Ne) = 20.18 g/mol
Number of moles added (n₃):
n₃ = mass / molar mass
n₃ = 25.0 g / 20.18 g/mol
Calculating the new volume (V₃):
V₃ = [(n₁ + n₃) * R * T] / P
V₃ = [(1.50 mol + (25.0 g / 20.18 g/mol)) * R * T] / P
Since the pressure (P) and temperature (T) remain constant, we can simplify the equation:
V₃ = (V₁ * (n₁ + n₃)) / n₁
V₃ = (8.00 L * (1.50 mol + (25.0 g / 20.18 g/mol))) / 1.50 mol
Therefore, the new volume of the gas after adding 25.0 g of neon is (8.00 L * (1.50 mol + (25.0 g / 20.18 g/mol))) / 1.50 mol.
c. When 3.50 mol of neon is added, we can calculate the new volume (V₄) using the same equation as in scenario b:
V₄ = (V₁ * (n₁ + n₄)) / n₁
V₄ = (8.00 L * (1.50 mol + 3.50 mol)) / 1.50 mol
Therefore, the new volume of the gas after adding 3.50 mol of neon is (8.00 L * (1.50 mol + 3.50 mol)) / 1.50 mol.