Question

At 1578.3 mmHg and 286 K, a skin diver exhales a 150.−mL bubble of air that is

77.0% N₂, 17.0% O₂, and 6.0% CO₂ by volume.

(a) What would the volume of the bubble be in mL if it were exhaled at the surface at

1.00 atm

and 298 K?

mL?

(b) How many moles of N₂ are in the bubble?

mol N_2?

Answer 1

a) P1V1/T1 = P2V2/T2

P1 = 1578.3 mmhg = 2.077 atm

V1 = 150 ml

T1 = 286 K

P2 = 1 atm

V2 = ?

T2 = 298

(2.077*150/286) = (1*V2/298)

so that, volume of bubble(V2) = 324.62 ml

b) total no of mol of air(n) = PV/RT

                             = (2.077*0.15)/(0.0821*286)

                             = 0.0133 mol

   no of mol of N2 are in the bubble = 0.0133*77/100

                                     = 0.01024 mol