Question

Samples of Neon and helium are placed into separate containers connected by a pinched rubber tube as shown. The sample of Ne is 5.00 L at 654 mmHg. _________

The sample of He is 3.00 L at 522 mmHg. When the clamp is removed and the gases are allowed to mix, what is the final pressure of the neon? The volume of the tube is negligible. Hint: each has their own initial volume and pressure, and when they are allowed to mix, the volume is the total.

Answer 1

[AI-generated answer — Bot GPT 3.5]

First, we need to find the total volume of the container when the gases are allowed to mix. Since the volume of the tube is negligible, the total volume will be the sum of the initial volumes of Ne and He:

Total volume = 5.00 L + 3.00 L = 8.00 L

Next, we can use the combined gas law to find the final pressure of Ne:

(P1)(V1) = (P2)(V2)

P1 = 654 mmHg
V1 = 5.00 L
V2 = 8.00 L (total volume)
P2 = final pressure of Ne

Plugging in the values:

(654 mmHg)(5.00 L) = (P2)(8.00 L)

3270 mmHg*L = 8P2

P2 = (3270 mmHg*L) / 8 L
P2 = 408.75 mmHg

Therefore, the final pressure of Ne is 408.75 mmHg.