Respuesta :
Answer:
- Question 1: You cannot make sure the ballon will not burst before it reaches 3,000m altitude because the volume will exceed 47.0 liters
- Question 2: 5.0 g
Explanation:
1. Data
- Maximum volume of the balloon: 47.0 liters
At 3,000 m:
- Temperature at 3,000 m, Tâ: 6.0ÂșC
- Pressure at 3,000m, Pâ: 0.565atm
- Volume, Vâ = ?
At ground:
- Temperature at ground, Tâ: 20ÂșC
- Pressure at ground, Pâ: 765mmHg
- Volume at ground, Vâ: 30.0 liters
Questions:
- Will the balloon's volume be more than 47.0 liter?
- Amount in grams of helium in the ballon:?
2. Unit conversions:
a) Convert Pâ to atm:
- Pâ = 765mmHg Ă 1 atm/760.0mmHg = 1.00657895atm
b) Convert the temperatures to kelvin:
- Tâ = 6.0 + 273.15 = 279.15K
- Tâ = 20 + 273.15 = 293.15K
3. Solution
Question 1.
Calculate the final volume of the balloon:
Since the number of moles of helium gas inside the balloon remains unchanged, pV = nRT yields the combined gas law:
     [tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}[/tex]
From which you can solve for Vâ and compute it:
    [tex]V_1=\dfrac{P_2V_2T_1}{P_1T_2}[/tex]
    [tex]V_1=\dfrac{1.00657895atm\cdot 30.0liter\cdot 279.15K}{0.565atm\cdot 293.15K}[/tex]
    [tex]V_1\approx51liters[/tex]
Then, the volume of the balloon will exceed the gurantee of the manufacturer.
Question 2.
a) Calculate the number of moles of helium
  [tex]pV=nRT[/tex]
  [tex]n=\dfrac{pV}{RT}[/tex]
  [tex]n=\dfrac{1.00657895atm\times 30.0liter}{0.08206atm\cdot liter/(K\cdot mol)\times 293.15K}[/tex]
  [tex]n=1.2558mol[/tex]
b) Calculate the mass in grams of helium
Use the atomic mass of helium: 4.003g/mol
- mass = number of moles Ă molar mass
- mass = 1.2558mol Ă 4.003g/mol = 5.02g
Rounding to 2 significant figures: 5.0g