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Answer:
2.148kg
Explanation:
To calculate the mass of air in the tire, we'll use the ideal gas law, which relates pressure, volume, temperature, and the gas constant. Here’s how we can proceed:
Given data:
- Volume of the tire \( V = 0.6 \text{ m}^3 \)
- Gauge pressure \( P_{\text{gauge}} = 200 \text{ kPa} \)
- Temperature \( T = 20^\circ \text{C} = 20 + 273.15 \text{ K} = 293.15 \text{ K} \)
Firstly, convert the gauge pressure to absolute pressure:
\[ P_{\text{absolute}} = P_{\text{gauge}} + P_{\text{atm}} \]
Assuming standard atmospheric pressure (\( P_{\text{atm}} \)) is approximately \( 101.325 \) kPa:
\[ P_{\text{absolute}} = 200 \text{ kPa} + 101.325 \text{ kPa} = 301.325 \text{ kPa} \]
Now, use the ideal gas law:
\[ PV = nRT \]
Where:
- \( P \) is the absolute pressure (in Pa),
- \( V \) is the volume of the tire (in m\(^3\)),
- \( n \) is the number of moles of gas,
- \( R \) is the universal gas constant (\( 8.314 \text{ J/(mol·K)} \)),
- \( T \) is the temperature in Kelvin.
First, calculate the number of moles of air \( n \):
\[ n = \frac{P V}{R T} \]
Convert the pressure \( P \) to Pa:
\[ P = 301.325 \times 10^3 \text{ Pa} \]
Now, substitute the values:
\[ n = \frac{301.325 \times 10^3 \text{ Pa} \times 0.6 \text{ m}^3}{8.314 \text{ J/(mol·K)} \times 293.15 \text{ K}} \]
Calculate \( n \):
\[ n \approx \frac{180.795 \times 10^3 \text{ m}^3\text{·Pa}}{2437.2 \text{ J/K}} \]
\[ n \approx 74.2 \text{ moles} \]
Finally, calculate the mass of air \( m \):
\[ m = n \times M \]
Where \( M \) is the molar mass of air (approximately \( 28.97 \text{ g/mol} \)):
\[ m \approx 74.2 \text{ moles} \times 28.97 \text{ g/mol} \]
\[ m \approx 2148 \text{ g} \]
Therefore, the mass of air in the tire is approximately \( 2148 \) grams, or \( 2.148 \) kilograms.