Three samples of gas each exert 740 mm hg in separate 2 l containers what pressure do they exert if they are all placed in a single 4l container?

P(total) = 1110 mmHg

Explanation:

According to the Dalton law of partial pressure,

The pressure exerted by mixture of gases are equal to the sum of partial pressure of individual gases.

P(total) = P1 + P2 + P3+ + Pn

Given data:

Sample A = 740 mmHg

Sample B = 740 mmHg

Sample C = 740 mmHg

Total pressure = ?

Solution:

Sample A:

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

P₂ = 740 mmHg × 2L/4L

P₂ = 370 mmHg

Sample B:

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

P₂ = 740 mmHg × 2L/4L

P₂ = 370 mmHg

Sample C:

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

P₂ = 740 mmHg × 2L/4L

P₂ = 370 mmHg

Total pressure:

P(total) = P1 + P2 + P3

P(total) =  370 mmHg +  370 mmHg+  370 mmHg

P(total) = 1110 mmHg

i believe the answer is true.