A. ) A goodyear blimp typically contains 5400 m cubed of helium at an absolute pressure of 1.1*10^5 Pa. The temperature of the helium is 280 K. What is the mass (in kg) of the helium of the blimp?
B. ) Estimate the spacing between the centers of neighboring atoms in a piece of solid aluminum, based on knowledge of the density (2700 kg/m^3) and atommic mass (26.9815 u) of aluminum.
a) use PV=nRT to find n, the number of moles of He. The Molar mass of He=4 g mol^-1. Hence the mass (g) is 4xn. then convert to kg.
the mass of aluminum cup
A. To find the mass of the helium in the blimp, we can use the ideal gas law equation:
PV = nRT
Where:
P = absolute pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
First, we need to convert the given volume from cubic meters to liters:
V = 5400 m^3 * 1000 L/m^3 = 5.4 * 10^6 L
Next, we can rearrange the equation to solve for the number of moles:
n = PV / RT
Now, we need to convert the pressure from Pascals to atmospheres:
P = 1.1 * 10^5 Pa * (1 atm / 101325 Pa) = 1.0841 atm
The ideal gas constant, R, is equal to 0.0821 L·atm/(mol·K).
Finally, we can plug the values into the equation and solve for n:
n = (1.0841 atm) * (5.4 * 10^6 L) / (0.0821 L·atm/(mol·K) * 280 K)
By canceling out the units, we find that the number of moles is approximately:
n ≈ 2304.52 mol
To find the mass of the helium, we can use the molar mass of helium (4.0026 g/mol):
Mass = n * molar mass
Mass ≈ 2304.52 mol * 4.0026 g/mol = 9219.831 kg
Therefore, the mass of the helium in the blimp is approximately 9219.831 kg.
B. The spacing between the centers of neighboring atoms in a solid can be estimated using the density and atomic mass. The formula for the volume of a single aluminum atom is given by:
V = (molar mass / density) * (1 / Avogadro's number)
First, we need to convert the atomic mass from atomic mass units (u) to kilograms:
Atomic mass = 26.9815 u * (1.66054 * 10^(-27) kg/u) = 4.4802561 * 10^(-26) kg
Next, we can use the formula to calculate the volume of a single atom:
V = (4.4802561 * 10^(-26) kg / 2700 kg/m^3) * (1 / 6.022 * 10^23 atoms/mol)
By canceling out units, we find that the volume of a single atom is approximately:
V ≈ 8.373681 * 10^(-29) m^3
Now, let's assume the shape of the aluminum atoms is spherical and calculate the radius:
V = (4/3) * π * r^3
8.373681 * 10^(-29) m^3 = (4/3) * π * r^3
Solving for r, the radius of an atom, we find:
r ≈ (3 * 8.373681 * 10^(-29) m^3 / (4 * π))^(1/3)
The spacing between the centers of neighboring atoms would be twice the radius, so:
Spacing ≈ 2 * r
By plugging the calculated values into the equation, we find that the spacing between the centers of neighboring atoms in solid aluminum is approximately:
Spacing ≈ 2 * [(3 * 8.373681 * 10^(-29) m^3 / (4 * π))^(1/3)] meters.
A. To find the mass of helium in the blimp, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, we convert the volume from cubic meters to liters (1 cubic meter = 1000 liters):
Volume (V) = 5400 m^3 * 1000 liters/m^3 = 5,400,000 liters
Next, we need to convert the pressure from Pascal (Pa) to atmospheres (atm) because the ideal gas constant (R) is usually expressed in atm:
Pressure (P) = 1.1 * 10^5 Pa * (1 atm / 101325 Pa) ≈ 1.0844 atm
Using the ideal gas law equation, we can solve for the number of moles (n):
n = PV / RT
Rearranging the equation to solve for mass (m):
m = n * Molar mass
The molar mass of helium is approximately 4.0026 grams/mole.
Now we can substitute the values into the equation:
m = n * 4.0026 g/mol
To find n, we can rearrange the ideal gas law equation again:
n = PV / RT
Substituting the values:
n = (1.0844 atm) * (5,400,000 liters) / ((0.0821 L * atm / K * mol) * 280 K)
Now we can calculate n:
n ≈ 227.54 moles
Finally, substitute n back into the equation to find the mass:
m = 227.54 moles * 4.0026 g/mol ≈ 910.35 grams
Therefore, the mass of the helium in the Goodyear blimp is approximately 910.35 grams, or 0.91035 kg.
B. In order to estimate the spacing between the centers of neighboring atoms in solid aluminum, we need to use the density (ρ) and atomic mass (m) of aluminum.
The density of aluminum is given as 2700 kg/m^3. This represents the mass of aluminum per unit volume.
The molar mass (atomic mass) of aluminum is given as 26.9815 u (atomic mass units).
To find the spacing between the centers of neighboring atoms, we can use the equation:
Spacing = (Volume of one atom)^(1/3)
The volume of one aluminum atom can be calculated using the formula:
Volume of one atom = (Molar mass / Density) * (1 / Avogadro's number)
Avogadro's number is approximately 6.022 × 10^23 atoms/mol.
First, convert the molar mass of aluminum from atomic mass units (u) to grams/mol:
molar mass (M) = 26.9815 u * (1.6605 × 10^(-27) kg/u) * (1000 g/kg) ≈ 0.045 g/mol
Now, substitute the values into the equation to find the volume of one atom:
Volume of one atom = (0.045 g/mol / 2700 kg/m^3) * (1 / (6.022 × 10^23 atoms/mol))
Volume of one atom ≈ 2.488 × 10^(-29) m^3
Finally, calculate the spacing between the centers of neighboring atoms:
Spacing = (2.488 × 10^(-29) m^3)^(1/3)
Spacing ≈ 2.683 × 10^(-10) meters
Therefore, the estimated spacing between the centers of neighboring atoms in solid aluminum is approximately 2.683 × 10^(-10) meters.