QuestionApril 29, 2025

A bicycle tire has a pressure of 6.43times 10^5N/m^2 at a temperature of 18 {}^circ C and contains 1.98 L of gas. What will its pressure be, if you let out an amount of air that has a volume of 142cm^3 at atmospheric pressure? Assume tire temperature and volume remain constant. square Pa

A bicycle tire has a pressure of 6.43times 10^5N/m^2 at a temperature of 18 {}^circ C and contains 1.98 L of gas. What will its pressure be, if you let out an amount of air that has a volume of 142cm^3 at atmospheric pressure? Assume tire temperature and volume remain constant. square Pa
A bicycle tire has a pressure of 6.43times 10^5N/m^2 at a temperature of 18 {}^circ C and contains 1.98 L of gas. What will its pressure be, if you let out an amount of air that has a volume of 142cm^3 at atmospheric pressure? Assume tire temperature and volume remain constant. square Pa

Solution
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Answer

6.93 \times 10^5 \, \text{Pa} Explanation 1. Convert volumes to consistent units Initial volume V_1 = 1.98 \, \text{L} = 1980 \, \text{cm}^3. Volume of air let out V_{\text{out}} = 142 \, \text{cm}^3. 2. Calculate remaining volume Remaining volume V_2 = V_1 - V_{\text{out}} = 1980 \, \text{cm}^3 - 142 \, \text{cm}^3 = 1838 \, \text{cm}^3. 3. Apply Boyle's Law **Boyle's Law**: P_1 V_1 = P_2 V_2. Solve for P_2: \[ P_2 = \frac{P_1 V_1}{V_2} \] 4. Substitute values and calculate P_2 = \frac{6.43 \times 10^5 \, \text{N/m}^2 \times 1980 \, \text{cm}^3}{1838 \, \text{cm}^3} P_2 = \frac{6.43 \times 10^5 \times 1980}{1838} \, \text{N/m}^2 P_2 \approx 6.93 \times 10^5 \, \text{N/m}^2

Explanation

1. Convert volumes to consistent units<br /> Initial volume $V_1 = 1.98 \, \text{L} = 1980 \, \text{cm}^3$. Volume of air let out $V_{\text{out}} = 142 \, \text{cm}^3$.<br /><br />2. Calculate remaining volume<br /> Remaining volume $V_2 = V_1 - V_{\text{out}} = 1980 \, \text{cm}^3 - 142 \, \text{cm}^3 = 1838 \, \text{cm}^3$.<br /><br />3. Apply Boyle's Law<br /> **Boyle's Law**: $P_1 V_1 = P_2 V_2$. Solve for $P_2$: <br />\[ P_2 = \frac{P_1 V_1}{V_2} \]<br /><br />4. Substitute values and calculate<br /> $P_2 = \frac{6.43 \times 10^5 \, \text{N/m}^2 \times 1980 \, \text{cm}^3}{1838 \, \text{cm}^3}$<br /><br /> $P_2 = \frac{6.43 \times 10^5 \times 1980}{1838} \, \text{N/m}^2$<br /><br /> $P_2 \approx 6.93 \times 10^5 \, \text{N/m}^2$
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