Answer to Question #83513 in General Chemistry for Trevor

Question

Container A holds 782 mL of ideal gas at 2.80 atm. Container B holds 174 mL of ideal gas at 4.40 atm. If the gases are allowed to mix together, what is the resulting pressure?

Answer 1

Answer on Question #83513

Container A holds 782 mL of ideal gas at 2.80 atm. Container B holds 174 mL of ideal gas at 4.40 atm. If the gases are allowed to mix together, what is the resulting pressure?

Solution

We should use Dalton's Law of Partial Pressures to calculate the resulting pressure:

P_total = P_A + P_B

Find pressure (P_A) of ideal gas that was in the container A when two gases were mixed together. We should use Boyle's Law:

P₁V₁ = P₂V₂

where P₁ = 2.80 atm, V₁ = 782 mL, P₂ = P_{A ,} V₂ = 782 mL + 174 mL = 956 mL

2.80*782 = P_A*956

P_A = 2.29 atm

Find pressure (P_B) of ideal gas that was in the container B when two gases were mixed together. We should use Boyle's Law:

P₁V₁ = P₂V₂

where P₁ = 4.40 atm, V₁ = 174 mL, P₂ = P_{B ,} V₂ = 782 mL + 174 mL = 956 mL

4.40*174 = P_B*956

P_B = 0.8 atm

P_total = P_A + P_B= 2.29 atm + 0.8 atm = 3.09 atm

Answer: 3.09 atm