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A 6.35 L sample of carbon monoxide is collected at 55.0◦C and 0.892 atm. What volume will the gas occupy at 1.05 atm and 59.0◦C? 1. 1.96 L 2. 6.68 L 3. 5.46 L 4. 6.10 L 5. 4.82 L

Respuesta :

Edufirst

Answer: option 3. 5.46 L

Explanation:

1) Data

  • V₁ = 6.35 L
  • T₁ = 55.0°C
  • P₁ = 0.892 atm
  • V₂ = ?
  • T₂ = 59.0°C
  • P₂ = 1.05 atm

2) Chemical principle

  • Combined law for ideal gases:

        [tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

3) Solution

a) Conversion of units

The ideal gas law is valid for temperature in absolute units (Kelvin); hence, you have to converte °C to K.

  • T₁ = 55.0 + 273.15 K = 328.15 K
  • T₂ = 59.0 + 273.15 K = 332.15 K

b) Clear V₂ from the combined law of gases

[tex]V_2 =\frac{P_1V_1T_2}{P_2T_1}[/tex]

c) Substitute values and compute

[tex]V_2=\frac{0.892atm6.35L332.15K}{1.05atm328.15K}=5.46L[/tex]

The answer must be reported with three significant figures, which is the number used for the data.

Answer: the gas will ocupy 5.46 liter at the new condictions of pressure and temperature.

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