The process by which a radioactive isotope loses protons or other materials from its nucleus is called

Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21 ∘C∘C. Express the pressure

Volgvan

The question is incomplete, complete question is ;

A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of $O_2$ and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.

Answer:

Partial pressure of the oxygen gas is 3.91 atm.

Partial pressure of the helium gas is 20.0 atm

Total pressure of the gases is 24.0 atm

Explanation:

Moles of oxygen gas = $n_1=\frac{51.8}{32 g/mol}=1.619 mol$

Moles of helium gas = $n_2=\frac{33.1 g}{4 g/mol}=8.275 mol$

Total moles of gas = $n_1+n_2=(1.619 +8.275 ) mole=9.894 mol$

Volume of the cylinder = V = 10.0 L

Total pressure in the cylinder = P = ?

Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K

PV = nRT ( ideal gas equation )

$P=\frac{nRT}{V}$

$=\frac{9.894 mol\times 0.0821 atm L/mol K\times 294 K}{10.0 L}$

P = 23.88 atm ≈ 23.9

Partial pressure of the individual gas will be determined by the help of Dalton's law:

partial pressure = Total pressure × mole fraction of gas

Partial pressure of the oxygen gas

$p_{1}=P\times \chi_{1}=P\times \frac{n_1}{n_1+n_2}$

$p_1=23.88 atm\times \frac{1.619 mol}{9.894 mol}=3.91 atm$

Partial pressure of the helium gas

$p_{2}=P\times \chi_{2}=P\times \frac{n_2}{n_1+n_2}$

$p_2=23.88 atm\times \frac{8.275 mol}{9.894 mol}=19.97 atm\approx 20.0 atm$

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