Question

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?

Answer 1

Pressure of gas mixture : P tot = 46,588 atm

Further explanation

Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases

Can be formulated:

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

The partial pressure is the pressure of each gas in a mixture

Since the gas mixture consists of He and CO₂, we determine the pressure of each gas then we add them together to get the total pressure

We assume the ideal gas, so we use the formula:

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm, N / m²

V = volume, liter

n = number of moles

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

T = temperature, Kelvin

n = N / No

n = mole

No = Avogadro number (6.02.10²³)

n = m / m

m = mass

M = relative molecular mass

 

• 1. He

M = 4 g / mol

m = 48.5 g

n = m: M

n = 48.5: 4

n = 12,125 mole

PV = nRT

$\rm P=\dfrac{nRT}{V}\\\\P=\dfrac{12.125\times 0.082\times 398}{10}\\\\P=\boxed{39,571\:atm}}$

2. CO₂

M = 44 g / mol

m = 94.6 g

n = m: M

n = 94.6: 44

n = 2.15 mole

PV = nRT

$\rm P=\dfrac{nRT}{V}\\\\P=\dfrac{2.15\times 0.082\times 398}{10}\\\\P=\boxed{7.0167\:atm}}$

P tot = P He + P CO₂

P tot = 39,571 + 7.0167

P tot = 46,588 atm

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Answer 2

Answer:

• 46.6 atm

Explanation:

1) Data:

a) V = 10.0 l

b) He, m₁ = 48.5 g

c) CO₂, m₂ = 94.6 g

d) T = 398 K

e) p = ?

2) Formulae:

a) ideal gas equation: pV = nRT

b) Number of moles, n: n = mass in grams / molar mass

3) Solution:

a) Number of moles of He:

• atomic mass of He: 4.003 g/mol

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

b) Number of moles of CO₂:

• molar mass of CO₂: 44.01 g/mol

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

c) Total number of moles of the gas mixture:

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

d) Pressure, p:

• pV = nRT ⇒ p = nRT / V

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