What pressure would a gas mixture in a 10.0 l tank exert if it were composed of 48.5 g he and 94.6 g co2 at 398 k?

Gas mixture pressure: P tot = 46.588 atm

Additional information

Dalton's law of partial pressures indicates that the aggregate pressure of a gas mixture equals the total of the individual gases' partial pressures.

This can be expressed as:

P tot = P1 + P2 + P3....

Each gas's partial pressure represents its pressure within the mixture.

We calculate the pressure of He and CO₂ from the mixture and sum them to obtain the overall pressure. We apply the ideal gas equation:

Where:

P = pressure, atm, N / m²

V = volume, in liters

n = mole count

R = gas constant = 0.082 l.atm / mol K (for P = atm, V = liters) or 8.314 J / mol K (for P = Pa or N / m², V = m³)

T = temperature in Kelvin

n = N / No

n = moles

No = Avogadro's number (6.02 x 10²³)

n = m / M

m = mass

M = molar mass

• 1. He

M = 4 g / mol

m = 48.5 g

n = m / M

n = 48.5 / 4

n = 12.125 moles

PV = nRT

2. CO₂

M = 44 g / mol

m = 94.6 g

n = m / M

n = 94.6 / 44

n = 2.15 moles

PV = nRT

P tot = P He + P CO₂

P tot = 39.571 + 7.0167

P tot = 46.588 atm

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Which equation aligns with the ideal gas law

Which law relates to the ideal gas law

Answer:

• 46.6 atm

Explanation:

1) Information:

a) Volume, V = 10.0 l

b) Helium, m₁ = 48.5 g

c) Carbon dioxide, m₂ = 94.6 g

d) Temperature, T = 398 K

e) Pressure, p =?

2) Equations:

a) ideal gas law: pV = nRT

b) Moles, n: n = mass in grams / molar mass

3) Calculation:

a) Calculating moles of He:

• Atomic weight of He: 4.003 g/mol

• n₁ = 48.5 g / 4.003 g/mol = 12.12 mol

b) Calculating moles of CO₂:

• Molar weight of CO₂: 44.01 g/mol

• n₂ = 94.6 g / 44.01 g/mol = 2.15 mol

c) Total moles in the mixture:

• n = n₁ + n₂ = 12.12 mol + 2.15 mol = 14.27 mol

d) Finding Pressure, p:

• pV = nRT ⇒ p = nRT / V

• p = 14.27 mol × 0.0821 atm-l/K-mol × 398K / 10.0l = 46.6 atm