Q)The molecular mass of
helium is 4 g/mol, the Boltzmann’s constant is 1.38066 ×
10−23 J/K, the universal gas constant is 8.31451 J/K · mol, and Avogadro’s number
is 6.02214 × 1023 1/mol. Given: 1 atm = 101300 Pa.
How many atoms of helium gas are required to fill a balloon to diameter 30 cm at 12◦C and 0.95 atm?
10−23 J/K, the universal gas constant is 8.31451 J/K · mol, and Avogadro’s number
is 6.02214 × 1023 1/mol. Given: 1 atm = 101300 Pa.
How many atoms of helium gas are required to fill a balloon to diameter 30 cm at 12◦C and 0.95 atm?
Sol:
PV = nRT
n = PV/RT
Number of atoms (N)= Avogadro's number (A) * n
N = A*n = A*(PV/RT)
P = 0.95 * 101300 Pa
P = 96235 Pa
V = (4/3) pi R^3
R = 0.15 m
V = (4/3) pi (0.15 m)^3
V = 0.01414 m^3
T = 273.15 + 12
T = 285.15 K
N = 6.02214 x 10^23 atoms/mole * (96235 Pa * 0.01414 m^3) / (8.31451 J/K-mol * 285.15 K)
N = 3.456 x 10^23 atoms
n = PV/RT
Number of atoms (N)= Avogadro's number (A) * n
N = A*n = A*(PV/RT)
P = 0.95 * 101300 Pa
P = 96235 Pa
V = (4/3) pi R^3
R = 0.15 m
V = (4/3) pi (0.15 m)^3
V = 0.01414 m^3
T = 273.15 + 12
T = 285.15 K
N = 6.02214 x 10^23 atoms/mole * (96235 Pa * 0.01414 m^3) / (8.31451 J/K-mol * 285.15 K)
N = 3.456 x 10^23 atoms
