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Crank
3 years ago
5

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?
Chemistry
1 answer:
iren [92.7K]3 years ago
6 0

Answer:

The resulting pressure is 2.81 atm

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

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If have a volume of 18 L of a gas at a temperature of 272 K and a pressure of 90 atm, what will be the pressure of the gas if ra
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Answer:

P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)

Explanation:

Given:

P₁ = 90 atm                    P₂ = ?

V₁ = 18 Liters(L)              L₂ = 12 Liters(L)      

=> decrease volume => increase pressure

=> volume ratio that will increase 90 atm is (18L/12L)                                                                  

T₁ = 272 Kelvin(K)          T₂ = 274 Kelvin(K)

=>  increase temperature => increase pressure

=> temperature ratio that will increase 90 atm is (274K/272K)

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P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)

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