When 6.0 L of He(g) and 10. L of N2(g), both at 0oC and 1.0 atm, are pumped into an evacuated 4.0 L rigid container, the final pressure in the container at 0o C is :

(A) 2.0 atm
(B) 4.0 atm
(C) 6.4 atm
(D) 8.8 atm
(E) 16 atm

4 atm. Option B.

Explanation:

We use the Ideal Gases Law to solve this problem: P . V = n . R .T

We need the moles of each gas:

6L . 1atm = n . 0.082 . 273K → (6L . 1atm)/(0.082 . 273K)= n → 0.27 mol He

10L . atm = n . 0.082 . 273K → (10L. 1atm)/(0.082 . 273K) = n → 0.45 mol N₂

Total moles on the mixture → 0.27 mol He + 0.45 mol N₂ = 0.72 moles

Now, we apply the Ideal Gases Law again.

Pressure . 4L = 0.72 moles . 0.082 . 273K

P = (0.72 moles . 0.082 . 273K) 4L → 4.02 atm

The final pressure is 4.0 atm (option B)

Explanation:

Step 1: Data given

Volume He = 6.0 L

Volume N2 = 10.0 L

Temp = 0 °C

Pressure = 1.0 atm

new total volume = 4.0 L

Step 2: Calculate new pressure of He

P1*V1 = P2V2

1.0 * 6.0 = P2 * 4.0

P2 = 6.0 / 4.0 atm

Step 3: Calculate new pressure of N2

P1*V1 = P2V2

1.0 * 10.0 = P2 * 4.0

P2 = 10.0 / 4.0 atm

Step 4: Calculate total pressure

P = 6/4 atm + 10/4 atm = 16/4 atm = 4.0 atm

The final pressure is 4.0 atm (option B)

answer: c.

explanation: