Scuba Diving Locations

This mixture contains 1.96 g He and 60.8 g O2 confided in a 5.00 L tank at 25.0 degrees celsius. What are:

A) the mole fractions of two gases?
B) their partial pressures?
C) total pressure?

find moles, using molar masses:
1.96 g He @ 4.00 g/mole = 0.49 moles Helium

60.8 g O2 @ 32.0 g/mol = 1.90 moles oxygen

total moles = 2.39 moles
===========================

A) the mole fractions of two gases?

Helium:
0.49 moles Helium / 2.39 total moles = 0.205 helium’s mole fraction

1.90 mole O2 / 2.39 total moles = 0.795 oxygen’s mole fraction

=================================

B) their partial pressures?

first find total pressure:
PV=nRT
P(5.00L) = 2.39 moles (0.08206 Litre-atm / mol-K)(298K)
total pressure = 11.69 atm

B) their partial pressures?
Helium:
11.69 (0.205 helium’s mole fraction) = 2.396 atm Helium
your answer(3sigfigs): 2.40 atm Helium partial pressure

Oxygen:
11.69 – 2.40 = 9.29 atm oxygen
your answer is: 9.29 atm Oxygen partial pressure
===============================
C) total pressure?

your answer(3sigfigs): 11.7 atm

I wouldn’t say I smoke heavily, probably 10 cigarettes a day for 4 years average, but I have been invited to do a scuba diving course in december and I was wondering whether this could harm me because of pressures and oxygen tanks or anything like that, any help or information is useful to me.

There shouldn’t be any issue unless you actually have a lung problem. I’m a pack a day smoker for 30 years and dive both commercially as well as recreationally. About the only issue some smokers have, is a tendency not to be able to equalize as easily as a non smoker. It’s due to the irritation smoke causes to nasal passages. No biggie though, you’ll still be able to equalize with practise.

answer correctly and u get 10 points

When people go scuba diving they take a gas tank full of gases. What gases are contained in the tank? Estimate the weight of gas present in a full scuba tank?

Volume = 11 L

P = 170 atm

People have used compressed air as their breathing medium since the advent of diving in the 1950s. Its main advantage is that it is readily available and relatively inexpensive to compress into cylinders. Nevertheless, air is not the "ideal" breathing mixture for diving. With a concentration of approximately 79% nitrogen, compressed air poses two potential problems for all divers: susceptibility to nitrogen narcosis (a condition resembling alcoholic intoxication) at deeper depths; and decompression sickness (DCS). Both of these can prove fatal to a diver. In an effort to reduce the ill effects of nitrogen on divers, nitrox was developed.

Nitrox is a generic term that can be used to describe any gaseous mixture of nitrogen and oxygen. In the context of technical diving, nitox is a mixture containing more oxygen than air. The two most commonly used nitrogen-oxygen mixtures contain 32% and 36% oxygen by volume. This differs significantly from compressed air, which contains approximately 21% oxygen by volume. While an increase of 12 to 16% oxygen by volume may not seem drastic, it allows divers to significantly extend their bottom time, and decreases their risk of developing DCS.

The term "mixed-gas diving" refers to any activity in which the diver breathes a mixture other than air or nitrox. The main incentive to dive with "non-air" gas mixtures is to avoid nitrogen narcosis. One type of mixed gas diving involves the use of heliox. This (helium 79% and oxygen 21%) mixture is often used for very deep diving. Unlike nitrogen, helium is not known to have an intoxicating effect at any depth; it has a lower density than nitrogen, making it easier to breathe; and in cases of extended submersion, it improves decompression.

Since there are verious compositions, I will solve the problem for taking helox as basis, that is (helium 79% and oxygen 21%) mixture.

In your question there is no information about the temperature. Therefore I will assume it constant.

To estimate the weight (mass) of the gas mixture in the tube, we have to know the total moles of the mixture. For this we can use the STP conditions to compare (represented by the subscript 1)
P1 = 1 atm
V1 = 22.4 L
n = 1 mol
T = 0°C = 273 K

Tube conditions (represented by the subscript 2)
P1 = 170 atm
V1 = 11 L
n = ? mol
T = 0°C = 273 K

P1V1 / n1T1 = P2V2 / n2T2 (cancel T1 and T2)

(1)(22.4) / 1 = (170)(11) / n2

n2 = 1870 / 22.4 = 83.5 mol

For gases volume percentages = mole percentages
Mole of He = 83.5 x 79/100 = 66.0 mol
Mole of O2 = 83.5 x 21/100 = 17.5 mol
———————————————————
TOTAL : ……………………………… 83.5 mol

Molar mass of He = 4 g/mol
Molar mass of O2 = 32 g/mol

Mass of He = 66.0 mol x 4 g/mol = 264 g
Mass of O2 = 17.5 mol x 32 g/mol = 560 g

Total mass = 824 g