A helium-filled balloon of the type used in long-distance flying contains 420,000 ft3 (1.2 times 107 L) of helium. Suppose you fill the balloon with helium on the ground, where the pressure is 737 mm Hg and the temperature is 16.0 °C. When the balloon ascends to a height of 2 miles, where the pressure is only 600. mm Hg and the temperature is −33 °C, what volume is occupied by the helium gas? Assume the pressure inside the balloon matches the external pressure.
To solve this problem, we can use the Combined Gas Law, which relates the initial and final temperatures, pressures, and volumes of a gas sample.
The Combined Gas Law equation is:
(P₁V₁) / (T₁) = (P₂V₂) / (T₂)
Where:
P₁ = initial pressure (in atmospheres)
P₂ = final pressure (in atmospheres)
V₁ = initial volume (in liters)
V₂ = final volume (in liters)
T₁ = initial temperature (in Kelvin)
T₂ = final temperature (in Kelvin)
Let's convert the given values to the appropriate units:
P₁ = 737 mm Hg = 737/760 atm (1 atm = 760 mm Hg)
P₂ = 600 mm Hg = 600/760 atm
V₁ = 1.2 × 10^7 L
T₁ = 16.0 °C = 16.0 + 273.15 K (converting to Kelvin)
T₂ = -33.0 °C = -33.0 + 273.15 K
Now we can substitute these values into the Combined Gas Law equation:
(P₁V₁) / (T₁) = (P₂V₂) / (T₂)
((737/760) × (1.2 × 10^7)) / (16.0 + 273.15) = ((600/760) × V₂) / (-33.0 + 273.15)
Now we can solve for V₂, the final volume occupied by the helium gas:
((737/760) × (1.2 × 10^7)) / (16.0 + 273.15) = ((600/760) × V₂) / (-33.0 + 273.15)
V₂ = ((737/760) × (1.2 × 10^7) × (-33.0 + 273.15)) / ((600/760))
Calculating this equation gives us the final volume occupied by the helium gas at a height of 2 miles.
To solve this problem, we will use the combined gas law, which relates the initial and final conditions of temperature, pressure, and volume. The combined gas law formula is:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Where:
P₁ and P₂ are the initial and final pressures
V₁ and V₂ are the initial and final volumes
T₁ and T₂ are the initial and final temperatures
Let's break down the information given in the problem:
1. Initial conditions:
- Pressure (P₁) = 737 mm Hg
- Volume (V₁) = 420,000 ft³
- Temperature (T₁) = 16.0 °C
2. Final conditions:
- Pressure (P₂) = 600 mm Hg
- Volume (V₂) = unknown (this is what we need to find)
- Temperature (T₂) = -33 °C
To use the combined gas law, we need to convert the temperatures to Kelvin. The Kelvin scale starts at absolute zero, so we need to add 273.15 to each Celsius temperature.
T₁ in Kelvin = 16.0 °C + 273.15 = 289.15 K
T₂ in Kelvin = -33 °C + 273.15 = 240.15 K
Now let's substitute the values into the combined gas law equation:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Solving for V₂:
V₂ = (P₂ * V₁ * T₂) / (P₁ * T₁)
Substituting the given values:
V₂ = (600 mm Hg * 420,000 ft³ * 240.15 K) / (737 mm Hg * 289.15 K)
To make the units consistent, we can choose either all metric units or all imperial units. Let's convert everything to metric units:
- Volume in ft³ to liters (1 ft³ ≈ 28.3168466 L)
V₂ = (600 mm Hg * 420,000 ft³ * 28.3168466 L) / (737 mm Hg * 289.15 K)
- Pressure in mm Hg to kPa (1 mm Hg ≈ 0.133322368 kPa)
V₂ = (600 mm Hg * 420,000 ft³ * 28.3168466 L) / (0.133322368 kPa * 289.15 K)
Calculate the value of V₂ to obtain the final volume occupied by the helium gas at the given conditions.