The pressure of a gas may be thought of as the force the gas exerts per unit area on the walls of it — Analytical Chemistry Chemistry Question
Snorkelling
The pressure of a gas may be thought of as the force the gas exerts per unit area on the walls of its container, or on an imaginary surface of unit area placed somewhere within the gas. The force arises from collisions between the particles in the gas and the surface. In an ideal gas, the collision frequency (number of collisions per second) with a surface of unit area is given by:
Z = p / (2π m k_B T)^(1/2)
where p is the pressure and T the temperature of the gas, m is the mass of the gas particles, and kB is the Boltzmann’s constant (kB = 1.38×10–23 J K–1).
At sea level, atmospheric pressure is generally around 101.3 kPa, and the average temperature on a typical British summer day is 15°C.
Using the approximation that air consists of 79 % nitrogen and 21 % oxygen, calculate the weighted average mass of a molecule in the air.
Model Answer
m = 0.79 × M(N2) + 0.21 × M(O2) = 0.79 × 28.02 g mol−1 + 0.21 × 32.00 g mol−1 = 28.86 g mol−1 = 4.79×10−26 kg
Human lungs have a surface area of approximately 75 m2. An average human breath takes around 5 seconds. Estimate the number of collisions with the surface of the lungs during a single breath on a typical British summer day. You should assume that the pressure in the lungs remains constant at atmospheric pressure; this is a reasonable approximation, as the pressure in the lungs changes by less than 1 % during each respiratory cycle.
Model Answer
Z = p / (2π m k_B T)^(1/2)
Number of collisions = Z × A × t = (101300 N m^-2 × 75 m^2 × 5 s) / (2π × 4.79×10^-26 kg × 1.3806×10^-23 J K^-1 × 287 K)^(1/2) = 1.47×10^28
The human lungs can operate against a pressure differential of up to one twentieth of atmospheric pressure. If a diver uses a snorkel for breathing, we can use this fact to determine how far below water the surface of the water she can swim.
The pressure experienced by the diver a distance d below the surface of the water is determined by the force per unit area exerted by the mass of water above her. The force exerted by gravity on a mass m is F = m g, where g = 9.8 m s–2 is the acceleration due to gravity.
Write down an expression for the mass of a volume of water with cross sectional area A and depth d.
Model Answer
m = ρ V = ρ A d
Derive an expression for the force exerted on the diver by the volume of water in 2.3, and hence an expression for the difference in pressure she experiences at depth d relative to the pressure at the water’s surface.
Model Answer
F = m g = ρ A d g
p = F / A = ρ d g
Calculate the maximum depth the diver can swim below the water surface, while still breathing successfully through a snorkel.
Model Answer
g = 9.8 m s−2
ρ = 1000 kg m−3
p_max = p_atm / 20 = 5065 Pa
d = p_max / (ρ g) = 5065 Pa / (1000 kg m−3 × 9.8 m s−2) = 0.52 m