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What mass of helium gas is present in a 13.8 L sample of this mixture at 265 K

What mass of helium gas is present in a 13.8 L sample of this mixture at 265 K-example-1
User Demurgos
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We have a mixture of gases. The total pressure of the mixture of gases is 750 mmHg. The partial pressure of He is 213 mmHg.

The partial pressure of a gas in a mixture of gases is the pressure of that gas if it is alone and occupying the whole volume at the same temperature.

So we can use the ideal gas law to get the number of moles of He using the partial pressure.

P * V = n * R * T

Where P is the partial pressure in this case, V is the volume, R is the ideal gas constant, n is the number of moles of He and T is the temperature.

R = 0.082 atm*L/(mol*K)

V = 13.8 L

T = 265 K

P = 213 mmHg = 213 mmHg * 1 atm/(760 mmHg) = 0.280 atm

P = 0.280 atm

We can replace the given values and solve the equation for n.

P * V = n * R * T

n = P * V/(R * T)

n = 0.280 atm * 13.8 L /(0.082 atm*L/(mol*K) * 265 K)

n = 0.178 moles of He

Once we know the number of moles of He we can convert them into grams using the molar mass of He.

molar mass of He = 4.00 g/mol

mass of He = 0.178 moles of He * 4.00 g/mol

mass of He = 0.712 g

Answer: m = 0.712 g

User Monte Hurd
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