Chemistry, 03.01.2020 06:31 meababy2009ow9ewa

4.0 l of he(g), 6.0 l of n2(g), and 10. l of ar(g), all at 0°c and 1.0 atm, are pumped into an evacuated 8.0 l rigid container, the final pressure in the container at 0°c is

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Answer from: ricardorendon100

The final pressure in the container at 0°C is 2.49 atm

Explanation:

We apply the Ideal Gases law to know the global pressure.

We need to know, the moles of each:

P He . V He = moles of He . R . 273K

(1atm . 4L) / R . 273K = moles of He → 0.178 moles

P N₂ . V N₂ = moles of N₂ . R . 273K

(1atm . 6L) / R . 273K = moles of N₂ → 0.268 moles

P Ar . V Ar = moles of Ar . R . 273K

(1atm . 10L) / R . 273K = moles of Ar → 0.446 moles

Total moles: 0.892 moles

P . 8L = 0.892 mol . R . 273K

P = ( 0.892 . R . 273K) / 8L = 2.49 atm

R = 0.082 L.atm/mol.K

Answer from: naomi12360

The final pressure in the evacuation container at 0 °C is 2.499 atm

To solve this question, we'll begin by calculating the number of mole of each gas. This can be obtained as follow:

For He:

Volume (V) = 4 L

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

Number of mole (n) =? PV = nRT

1 × 4 = n × 0.0821 × 273

4 = n × 22.4133

Divide both side by 22.4133

n = 4 / 22.4133

n = 0.178 mole For N₂:

Volume (V) = 6 L

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

Number of mole (n) =? PV = nRT

1 × 6 = n × 0.0821 × 273

6 = n × 22.4133

Divide both side by 22.4133

n = 6 / 22.4133

n = 0.268 mole For Ar:

Volume (V) = 10 L

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

Number of mole (n) =?

PV = nRT

1 × 10 = n × 0.0821 × 273

10 = n × 22.4133

Divide both side by 22.4133

n = 10 / 22.4133

n = 0.446 mole

Next, we shall determine the total moles of the gas in the container.

Mole of He = 0.178 mole

Mole of N₂ = 0.268 mole

Mole of Ar = 0.446 mole

Total mole =?

Total mole = 0.178 + 0.268 + 0.446

Total mole = 0.892 mole

Finally, we shall determine the pressure in evacuation container. This can be obtained as follow:

Volume (V) = 8 L

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

Number of mole (n) = 0.892 mole

Pressure (P) =? PV = nRT

P × 8 = 0.892 × 0.0821 × 273

P × 8 = 19.993

Divide both side by 8

P = 19.993 / 8

P = 2.499 atm

Therefore, the final pressure in the evacuation container at 0 °C is 2.499 atm

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4.0 l of he(g), 6.0 l of n2(g), and 10. l of ar(g), all at 0°c and 1.0 atm, are pumped into an evacu...

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