Question

Container A holds 722 mL 722 mL of an ideal gas at 2.40 atm. 2.40 atm. Container B holds 169 mL 169 mL of a different ideal gas at 4.60 atm. 4.60 atm. If the gases are allowed to mix together, what is the resulting pressure?

Answer 1

The resulting pressure is 2.81 atm

Step-by-step explanation:

According to Dalton's Law of Partial Pressure, each of the gases (A and B) will exert their pressure independently. If we use Boyle's Law to calculate the pressure of each of the gases separately we have:

Pressure of gas A:

p1V1 = p2V2

p1 = 2.4 atm

V1 = 722 mL

V2 = 722 + 169 = 891 mL

p2 =?

Clearing p2:

p2 = (p1V1)/V2 = (2.4*722)/891 = 1.94 atm

Pressure of gas B:

p1 = 4.6 atm

V1 = 169 mL

V2 = 169+722 = 891 mL

p2=?

Clearing p:

p2 = (4.6*169)/891 = 0.87 atm

Dalton's expression for total partial pressures is equal to:

ptotal = pA + pB = 1.94+0.87 = 2.81 atm