To calculate the temperature of the air in the balloon using the given dimensions, we first need to find the volume of the balloon, assuming it is a sphere elongated by a factor of 23/14 in height.
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Determine the dimensions:
- Height (h) = 23 meters
- Width (diameter) (d) = 14 meters, therefore the radius (r) = d/2 = 14/2 = 7 meters.
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Calculate the volume of the balloon: Since the balloon can be approximated to an ellipsoid due to its elongated shape, the volume \( V \) of an ellipsoid is given by the formula:
\[ V = \frac{4}{3} \pi a b c \]
where \( a \), \( b \), and \( c \) are the semi-principal axes of the ellipsoid.
For our case:
- The height (along the vertical axis) \( c \) is half of 23 meters, which is \( c = \frac{23}{2} = 11.5 \) meters.
- The same radius applies for the width (along both horizontal axes), so:
- \( a = b = 7 \) meters.
Substituting the values into the volume formula: \[ V = \frac{4}{3} \pi (7)(7)(11.5) \]
Calculate:
\[ V = \frac{4}{3} \pi (49)(11.5) \approx \frac{4}{3} \pi (563.5) \approx 2362.35 \text{ cubic meters (approximately)} \]
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Using the Ideal Gas Law to find temperature: The ideal gas law is given by: \[ PV = nRT \] where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant (approximately \( 8.314 , \text{J/(mol K)} \)), and \( T \) is the temperature in Kelvin.
To find the temperature \( T \): \[ T = \frac{PV}{nR} \]
We need the pressure \( P \) and the number of moles \( n \) of air in the balloon. Assuming standard atmospheric pressure \( P \approx 101325 , \text{Pa} \) and that air is approximately \( 1 \text{ mole/L} \approx 0.001 \text{ m}^3 \), we can first approximate the number of moles.
- The density of air is about \( 1.225 , \text{kg/m}^3\), and the molar mass of air is about \( 28.97 , \text{g/mol} \approx 0.02897 \text{ kg/mol} \).
Let's find the mass of air: \[ m = \text{density} \times \text{volume} = 1.225 , \text{kg/m}^3 \times 2362.35 , \text{m}^3 \approx 2893.98 , \text{kg} \]
Now, calculating the number of moles \( n \): \[ n = \frac{m}{\text{molar mass}} = \frac{2893.98 , \text{kg}}{0.02897 , \text{kg/mol}} \approx 99996.67 , \text{moles} \]
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Now substitute into the ideal gas law: Given \( P = 101325 , \text{Pa} \), \( V \approx 2362.35 , \text{m}^3 \), and we already calculated \( n \):
\( R \) is approximately \( 8.314 , \text{J/(mol K)} \) (use proper units),
\[ T = \frac{PV}{nR} = \frac{101325 \times 2362.35}{99996.67 \times 8.314} \]
Calculate \( T \): \[ T \approx \frac{239317128.75}{833680.82} \approx 286.56 , \text{K} \]
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Convert Kelvin to Celsius: \[ T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \approx 286.56 - 273.15 \approx 13.41 , \text{°C} \]
Thus, the temperature of the air in the balloon is approximately \( 13.41 , \text{°C} \).