The equation to be used are:
PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles
The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.
PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa
n = 0.587 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.02024 mol
(101,183.9 Pa)V = (0.02024 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.931×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.931×10⁻⁴ m³ * 1000
V = 0.493 L
This is actually incorrect, the distance as stated above either from trough to trough or from crest to crest within a corresponding wave, would be referred or called as the wavelength of the given wave.
I think B may be the answer sorry if I'm wrong
C. They Typically Cantain AN -Oh group
