A Mylar Happy Birthday balloon contains 5.15 L of helium gas at 19.4oC. If the pressure remains constant, what will be the volume of helium in the balloon, in L, when the temperature is increased by 6.6oC?
(V1/T1) = (V2/T2)
Remember T must be in kelvin.
Why did the balloon go to school? Because it wanted to increase its air-telligence! 😄 Alright, let's calculate the new volume of helium in the balloon. We can use Charles's law, which states that the volume of a fixed amount of gas is directly proportional to its temperature, assuming pressure is constant.
So, we'll use the formula V₁ / T₁ = V₂ / T₂, where V₁ is the initial volume, T₁ is the initial temperature, V₂ is the final volume (what we want to find), and T₂ is the final temperature.
Given:
V₁ = 5.15 L
T₁ = 19.4°C
T₂ = 19.4°C + 6.6°C = 26°C
Using the formula, we can calculate V₂:
5.15 L / 19.4°C = V₂ / 26°C
Cross-multiplying:
V₂ = (5.15 L / 19.4°C) * 26°C
Calculating:
V₂ ≈ 6.89 L
So, when the temperature increases by 6.6°C, the volume of helium in the balloon will be approximately 6.89 L.
To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin when pressure and amount of gas are held constant.
First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15.
Initial temperature: T1 = 19.4oC + 273.15 = 292.55 K
Change in temperature: ΔT = 6.6oC
Next, we can set up the following equation:
(Volume 1) / (Temperature 1) = (Volume 2) / (Temperature 2)
Plugging in the given values:
(5.15 L) / (292.55 K) = (Volume 2) / (292.55 K + 6.6 K)
Now, we can solve for Volume 2:
(5.15 L) / (292.55 K) = (Volume 2) / (299.15 K)
Cross-multiplying:
(5.15 L) × (299.15 K) = (Volume 2) × (292.55 K)
Simplifying:
1542.1025 = (Volume 2) × (292.55 K)
Finally, we can solve for Volume 2:
Volume 2 = 1542.1025 / 292.55 K
Volume 2 ≈ 5.27 L
Therefore, when the temperature is increased by 6.6oC, the volume of helium in the balloon will be approximately 5.27 L.
To solve this problem, we can use the combined gas law, which states:
(P₁V₁/T₁) = (P₂V₂/T₂)
Where:
P₁ and P₂ are the initial and final pressures, respectively.
V₁ and V₂ are the initial and final volumes, respectively.
T₁ and T₂ are the initial and final temperatures, respectively.
Since the pressure remains constant, we can ignore it in the equation. Thus, the equation becomes:
(V₁/T₁) = (V₂/T₂)
Now, let's plug in the given values:
V₁ = 5.15 L (initial volume)
T₁ = 19.4°C (initial temperature)
T₂ = 19.4°C + 6.6°C = 26°C (final temperature, increased by 6.6°C)
Now we can solve for V₂ (final volume):
(V₁/T₁) = (V₂/T₂)
(5.15 L / 19.4°C) = (V₂ / 26°C)
To find V₂, we can rearrange the equation:
V₂ = (5.15 L * 26°C) / 19.4°C
V₂ ≈ 6.91 L
Therefore, when the temperature is increased by 6.6°C, the volume of helium in the balloon will be approximately 6.91 L.