Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that 1. The air pres

Question

Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that 1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose. 2. The temperature of the air is constant (body temperature). 3. The air acts as an ideal gas. 4. Salt water has an average density of around 1.03 g/cm3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm. What is the ratio of the molar concentration of gases in Gabor's lungs at the depth of 15 meters to that at the surface? (n/V) 15 m The molar concentration refers to e., the number of moles per unit volume. So you are asked to calculate (n/V) )surface Express your answer numerically to three significant digits. (n/V) 2.5 (m/V)surface Submit Hints My Answers Give Up Review Part Correct

Explanation / Answer

PV = n RT
P is pressure in N/m2
V is volume in m3
n is number of moles of gas in volume V
R is constant 8.31 J/mol - K
T is tempreture in Kelvin.
As person rises from 15 m depth to surface,
V, volumes of lungs and T, tempreture remian constant
Number of moles decrease due to decrease in pressure.
hence number of moles released = (P15m - Psurface) V / R T
(P15m - Psurface) = 1.5 Atm
Given that 1 Atm = 10 m of water of density 1.03 g/cm3
1.5 Atm = 15 m of water column = 15*1030*9.8 = 151410 N/m2   ( using P = h rho g)
V = 6x10-3 m3
T = 37 oC = 310 K
moles of air released = 151410* 6x10-3 / 8.31*310 = 0.35

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