Respuesta :
Answer:
13. 2.60 L.
14. 2.40 L.
Explanation:
We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If n and T are constant, and have different values of P and V:
(P₁V₁) = (P₂V₂)
13. A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the pressure is increased to 1.25 atm.
P₁ = 0.755 atm, V₁ = 4.31 L.
P₂= 1.25 atm, V₂ = ??? L.
∴ V₂ = (P₁V₁)/(P₂) = (0.755 atm)(4.31 L)/(1.25 atm) = 2.60 L.
14. 600.0 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at 2.00 at m.
P₁ = 8.0 atm, V₁ = 600.0 mL.
P₂= 2.0 atm, V₂ = ??? L.
∴ V₂ = (P₁V₁)/(P₂) = (8.0 atm)(600.0 mL)/(2.0 atm) = 2400/0 mL = 2.40 L.
13. The volume of the gas if the pressure is increased to 1.25 atm is 2.60 liters
14. The volume of the gas at 2.00 atm is 2400.0 mL
- For the first question
To determine the volume of the gas,
We will use the equation from Boyle's law
Boyle's law states that "the volume of a fixed mass of gas is inversely proportional to its pressure, provided that the temperature remains constant".
From Boyle's law, we have that
P₁V₁ = P₂V₂
Where P₁ is the initial pressure
V₁ is the initial volume
P₂ is the final pressure
and V₂ is the final volume
From the question
P₁ = 0.755 atm
V₁ = 4.31 liters
P₂ = 1.25 atm
Put the above parameters into the equation to determine the final volume,
That is,
P₁V₁ = P₂V₂ becomes
0.755 × 4.31 = 1.25 × V₂
∴ V₂ = [tex]\frac{0.755 \times 4.31}{1.25}[/tex]
V₂ = [tex]\frac{3.25405}{1.25}[/tex]
V₂ = 2.60324
V₂≅ 2.60 liters
Hence, the volume of the gas if the pressure is increased to 1.25 atm is 2.60 liters
- For the second question
We are to determine the volume of the gas at 2.00 atm
From the question
P₁ = 8.00 atm
V₁ = 600.0 mL
P₂ = 2.00 atm
Put the above parameters into the Boyle's law equation to determine the final volume,
That is,
P₁V₁ = P₂V₂ becomes
8 × 600 = 2.00 × V₂
∴ V₂ = [tex]\frac{8.00 \times 600.0}{2.00}[/tex]
V₂ = [tex]\frac{4800}{2.00}[/tex]
V₂ = 2400.0 mL
Hence, the volume of the gas at 2.00 atm is 2400.0 mL
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