A husband buys a helium-filled anniversary balloon for his wife. The balloon has a volume of 4.4 L in the warm store at 74 F. When he takes it outside, where the temperature is 55 F, he finds it has shrunk. By how much has the volume decreased?
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To calculate the decrease in volume of the helium-filled balloon when taken outside, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature. The equation for Charles's Law is:
V1 / T1 = V2 / T2
Where:
V1 = initial volume (in warm store) = 4.4 L
T1 = initial temperature (in warm store) = 74 F = 23.3 °C
V2 = final volume (outside) = ?
T2 = final temperature (outside) = 55 F = 12.8 °C
Let's substitute the values into the equation:
4.4 L / 23.3 °C = V2 / 12.8 °C
Now we can solve for V2:
V2 = (4.4 L / 23.3 °C) * 12.8 °C
V2 ≈ 2.42 L
Therefore, the volume of the balloon has decreased by approximately 2.42 L when taken outside.
To find the decrease in volume of the balloon when taken outside, you can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is constant.
Charles's Law can be expressed using the equation: V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature of the balloon, and V2 and T2 are the final volume and temperature.
In this case, the initial volume of the balloon is 4.4 L, and the initial temperature is 74 F. The final temperature is 55 F.
First, convert the temperatures to absolute scale (Kelvin) by adding 273.15 to each:
T1 = 74 F + 273.15 = 347.15 K
T2 = 55 F + 273.15 = 328.15 K
Next, use the equation V1/T1 = V2/T2 to find V2 (the final volume):
(V1/T1) / (V2/T2) = 1
V2 = V1 * (T2 / T1)
V2 = 4.4 L * (328.15 K / 347.15 K)
Calculate the final volume:
V2 = 4.4 L * (0.944)
The final volume, when the balloon is taken outside, is approximately 4.15 L.
To find the decrease in volume, subtract the final volume (4.15 L) from the initial volume (4.4 L):
Decrease in volume = 4.4 L - 4.15 L = 0.25 L
So, the volume of the balloon has decreased by 0.25 L when taken outside.