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An automobile tire volume of 0. 6m3 is inflated to a gauge pressure of 200kPa. Calculate the mass of air in the tire if the temperature is 20°C.

Sagot :

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.