Container A holds 717 mL of ideal gas at 2.50 bar . Container B holds 179 mL of ideal gas at 4.30 bar . If the gases are allowed

Question

Dmitrij [34]

4 years ago

14

Container A holds 717 mL of ideal gas at 2.50 bar . Container B holds 179 mL of ideal gas at 4.30 bar . If the gases are allowed

to mix together, what is the partial pressure of each gas in the total volume?

Chemistry

1 answer:

Nesterboy [21] · Answer 1 · 2021-12-30T10:34:46+03:00

Answer:

Pressure gas A

using boyles law

$p1}$ $V_{1}$ = $P_{2}$ V2

V2 = 717ml + 179 ml

= 896ml

∴ $P_{2}$ = 2.50 × 717ml/896ml

= 2.0 bar

Pressure B

P2 = 4.30 bar× 179ml/896ml

= 0.859bar

ptotal = $P_{A}$ + $P_{B}$

= 2.0 bar + 0.859 bar

= 2.859 bar

Explanation: Using Daltons law of partial pressure,the pressure is independently of each other when the gas is exerted.where we can use daltons law to find the pressure of each gas separately when it expands into the total volume in two containers.