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1.) A balloon contains 0.126 mol of gas and has a volume of 2.74 L .

If an additional 0.124 mol of gas is added to the balloon (at the same temperature and pressure), what will its final volume be?
Express your answer in liters to three significant figures.

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2.) A cylinder with a moveable piston contains 0.547 mol of gas and has a volume of 295 mL .

What will its volume be if an additional 0.342 mol of gas is added to the cylinder? (Assume constant temperature and pressure.)

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Edufirst

Answer:

  • Problem 1: 5.44L
  • Problem 2: 479mL

Explanation:

Problem 1.

a) Data:

  • n₁ = 0.126 mol
  • V₁ = 2.74 L
  • n₂ = 0.126mol + 0.124mol
  • V₂ = ?
  • Conditions: constant temperature and pressure

b) Physical principles:

The volume occupied by a gas, when the temperature and pressure remain unchanged, is proportional to the number of moles of gas.

          [tex]\dfrac{n_1}{V_1}=\dfrac{n_2}{V_2}[/tex]

c) Solution:

Solve for V₂ , substitute and compute:

        [tex]V_2=n_2\times\dfrac{V_1}{n_1}\\ \\\\ V_2=(0.126mol+0.124 mol)\times \dfrac{2.74L}{0.126mol}\\ \\ \\ V_2=5.44L[/tex]

Problem 2.

a) Data:

  • n₁ = 0.547mol
  • V₁ = 295 mL
  • V₂ = ?
  • n₂ = 0.547mol + 0.342 mol
  • Conditions: constant temperature and pressure

b) Physical principles:

The same as above: the volume occupied by a gas, when the temperature and pressure remain unchanged, is proportional to the number of moles of gas.

          [tex]\dfrac{n_1}{V_1}=\dfrac{n_2}{V_2}[/tex]

c) Solution:

Solve for V₂ , substitute and compute:

        [tex]V_2=n_2\times\dfrac{V_1}{n_1}\\ \\\\ V_2=(0.547mol+0.342 mol)\times \dfrac{295mL}{0.547mol}\\ \\ \\ V_2=479mL[/tex]

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