lisa asked in Science & MathematicsChemistry · 7 months ago

Charle's law ?

568 cm3 of chlorine at 25°C will occupy what volume at -25°C while the pressure remains constant?

A sample of nitrogen now occupies a volume of 250 mL at 25°C. What volume did it occupy at 95°C?

Oxygen gas is at a temperature of 40°C when it occupies a volume of 2.30 L. To what temperature should it be raised to occupy a volume of 6.50 L?

Hydrogen gas was cooled from 150°C to 50°C. Its new volume is 75.0 mL. What was its original volume?

A sample of neon gas at 50°C and with a volume of 2.50 L is cooled from 25°C. What was its original volume?

Fluorine gas at 300 K occupies a volume of 500 mL. To what temperature should it be lowered to bring the volume to 300 mL?

Helium occupies a volume of 3.8 L at –45°C. What was its initial temperature when it occupied 8.3 L?

A sample of argon gas is cooled and its volume went from 380 mL to 250 mL. If its final temperature was –55°C, what was its original temperature?

A container holds 50.0 mL of nitrogen at 25°C and at a constant pressure of 736 mm Hg. What will be its volume if the temperature increases by 35°C?

 

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  • 7 months ago
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    568 cm3 of chlorine at 25°C will occupy what volume at -25°C while the pressure remains constant?

    V₁/T₁ = V₂/T₂

    New volume, V₂ = V₁ × (T₂/T₁) = 568 × [(273 - 25)/(273 + 25)] cm³ = 473 cm³

    ====

    A sample of nitrogen now occupies a volume of 250 mL at 25°C. What volume did it occupy at 95°C?

    V₁/T₁ = V₂/T₂

    New volume, V₂ = V₁ × (T₂/T₁) = 250 × [(273 + 95)/(273 + 25)] mL = 309 mL

    ====

    Oxygen gas is at a temperature of 40°C when it occupies a volume of 2.30 L. To what temperature should it be raised to occupy a volume of 6.50 L?

    V₁/T₁ = V₂/T₂

    New temperature, T₂ = T₁ × (V₂/V₁) = (273 + 40) × (6.50/2.30) K = 885 K = 612°C

    ====

    Hydrogen gas was cooled from 150°C to 50°C. Its new volume is 75.0 mL. What was its original volume?

    V₁/T₁ = V₂/T₂

    Original volume, V₁ = V₂ × (T₁/T₂) = 75.0 × [(273 + 150)/(273 + 50)] mL = 98.2 mL

    ====

    A sample of neon gas at 50°C and with a volume of 2.50 L is cooled from 25°C. What was its original volume?

    V₁/T₁ = V₂/T₂

    Original volume, V₁ = V₂ × (T₁/T₂) = 2.50 × [(273 + 25)/(273 + 50)] L = 2.71 L

    (Note: The gas could NOT be cooled from 25°C to 50°C.)

    ====

    Fluorine gas at 300 K occupies a volume of 500 mL. To what temperature should it be lowered to bring the volume to 300 mL?

    V₁/T₁ = V₂/T₂

    New temperature, T₂ = T₁ × (V₂/V₁) = 300 × (300/500) K = 180 K

    ====

    Helium occupies a volume of 3.8 L at –45°C. What was its initial temperature when it occupied 8.3 L?

    V₁/T₁ = V₂/T₂

    Initial temperature, T₁ = T₂ × (V₁/V₂) = (273 - 45) × (8.3/3.8) K = 498 K = 225°C

    ====

    A sample of argon gas is cooled and its volume went from 380 mL to 250 mL. If its final temperature was –55°C, what was its original temperature?

    V₁/T₁ = V₂/T₂

    Original temperature, T₁ = T₂ × (V₁/V₂) = (273 - 55) × (380/250) = 331 K = 58°C

    ====

    A container holds 50.0 mL of nitrogen at 25°C and at a constant pressure of 736 mm Hg. What will be its volume if the temperature increases by 35°C?

    V₁/T₁ = V₂/T₂

    New volume, V₂ = V₁ × (T₂/T₁) = 50.0 × [(273 + 25 + 35)/(273 + 25)] mL = 55.9 mL

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