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

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

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:

Volume (V) = 4 L

Pressure (P) = 1 atm

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

Temperature (T) = 0 °C = 273 K

1 × 4 = n × 0.0821 × 273

4 = n × 22.4133

Divide both side by 22.4133

n = 4 / 22.4133

Volume (V) = 6 L

Pressure (P) = 1 atm

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

Temperature (T) = 0 °C = 273 K

1 × 6 = n × 0.0821 × 273

6 = n × 22.4133

Divide both side by 22.4133

n = 6 / 22.4133

Volume (V) = 10 L

Pressure (P) = 1 atm

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

Temperature (T) = 0 °C = 273 K

PV = nRT

1 × 10 = n × 0.0821 × 273

10 = n × 22.4133

Divide both side by 22.4133

n = 10 / 22.4133

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 = 0.178 + 0.268 + 0.446

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

P × 8 = 0.892 × 0.0821 × 273

P × 8 = 19.993

Divide both side by 8

P = 19.993 / 8

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

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answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.