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Mixtures of helium and oxygen are used in scuba diving tanks to help prevent "the bends". For a particular dive, 27.0 L of oxygen gas at 25.0°C and 0.900 atm, and 6.80 L of helium gas at 25.0°C and 1.80 atm were pumped into a tank with a volume of 5.70 L at 25 oC. Calculate the new partial pressure of helium (in atm).

User Hooney
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Final answer:

To find the new partial pressure of helium in the diving tank, calculate the moles of helium using the initial conditions and the ideal gas law, then use the moles, known volume of the tank, and temperature to calculate the final pressure. The new partial pressure of helium in the diving tank is 1.93 atm.

Step-by-step explanation:

To calculate the new partial pressure of helium (He) in a scuba tank after mixing with oxygen (O2), we first need to use the ideal gas law for both helium and oxygen before mixing. The ideal gas law is given by PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

Since the diver's tank is at 25.0°C, we can convert this temperature to Kelvin by adding 273.15, resulting in T = 298.15 K. We use the initial conditions for helium to find the moles of He:

nHe = (PHe * VHe) / (R * T)

Plugging in the given values:

nHe = (1.80 atm * 6.80 L) / (0.0821 L·atm/(mol·K) * 298.15 K) = 0.476 moles of He.

After the gases are mixed in the 5.70 L tank, the partial pressure of He is determined by rearranging the ideal gas law to solve for P:

PHe = (nHe * R * T) / Vtank

Substituting the known values:

PHe = (0.476 moles * 0.0821 L·atm/(mol·K) * 298.15 K) / 5.70 L = 1.93 atm.

Thus, the new partial pressure of helium in the diving tank is 1.93 atm.

User Peru
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