Problem 56
Question
A bicycle tire is filled with air to a pressure of \(75\) psi at a temperature of \(19^{\circ} \mathrm{C}\). Riding the bike on asphalt on a hot day increases the temperature of the tire to \(58^{\circ} \mathrm{C}\). The volume of the tire increases by \(4.0 \% .\) What is the new pressure in the bicycle tire?
Step-by-Step Solution
Verified Answer
The new pressure in the bicycle tire is approximately \(79.52\) psi.
1Step 1: Convert temperatures to Kelvin
First, we need to convert the given temperatures from Celsius to Kelvin, as the ideal gas law requires temperatures to be in Kelvin. We can do this using the formula \(T(K) = T(^\circ C) + 273.15\).
Initial temperature in Kelvin: \(T_1 = 19 + 273.15 = 292.15 K\)
Final temperature in Kelvin: \(T_2 = 58 + 273.15 = 331.15 K\)
2Step 2: Determine the final volume
The volume of the tire increases by \(4.0\%\). Denoting the initial volume as \(V_1\), the final volume can be calculated as \(V_2 = V_1 + 0.04 V_1 = 1.04V_1\).
3Step 3: Apply the ideal gas law formula
We will now apply the ideal gas law formula \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) to determine the final pressure.
We can rearrange the formula to solve for \(P_2\):
\(P_2 = \frac{P_1V_1T_2}{T_1V_2}\)
4Step 4: Substitute values and solve for the final pressure
Substitute given values, \(P_1 = 75 \text{ psi}\), \(T_1 = 292.15 K\), \(T_2= 331.15 K\), and \(V_2 = 1.04V_1\):
\(P_2 = \frac{75 \times V_1 \times 331.15}{292.15 \times 1.04V_1}\)
You can see that the volume terms \(V_1\) cancel out:
\(P_2 = \frac{75 \times 331.15}{292.15 \times 1.04}\)
Now, calculating the final pressure, \(P_2\):
\(P_2 \approx 79.52 \text{ psi}\)
The new pressure in the bicycle tire is approximately \(79.52\) psi.
Key Concepts
Temperature ConversionPressure CalculationVolume ChangeCelsius to Kelvin Conversion
Temperature Conversion
Understanding how to convert temperatures is essential in many scientific calculations, particularly when using the Ideal Gas Law. In our problem, the initial and final temperatures are given in Celsius, but they should be converted to Kelvin. The Kelvin scale is absolute and starts at absolute zero, which makes it the standard unit of temperature in scientific equations. Here's how you can perform the conversion:
This conversion ensures consistent calculations since Kelvin provides a true zero point, avoiding negative temperatures which Celsius can sometimes display.
- To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
This conversion ensures consistent calculations since Kelvin provides a true zero point, avoiding negative temperatures which Celsius can sometimes display.
Pressure Calculation
Pressure is a fundamental concept in physics and chemistry, especially when dealing with gases. In the context of our exercise, we use the Ideal Gas Law to determine the final pressure after changes in temperature and volume. The Ideal Gas Law is expressed as \( PV = nRT \), but in our scenario, we use the variation \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]This equation allows us to compare the initial and final states of the gas in the tire. To find the new pressure \( P_2 \), we rearrange it to isolate \( P_2 \). This formula considers how both temperature (increases) and volume (expands) affect pressure.
- Rearrange the formula to \( P_2 = \frac{P_1V_1T_2}{T_1V_2} \)
- Substitute existing values from the problem into this formula.
Volume Change
Volume plays an integral role in determining pressure changes for gases. In our problem, it is given that the volume increases by 4%. Volume change is a crucial factor in the Ideal Gas Law, especially under constant conditions like this scenario where the amount of gas doesn't change, but its volume does.To determine the final volume, \( V_2 \), when the volume increases:
- If initial volume is \( V_1 \), then \( V_2 = 1.04 \times V_1 \).
Celsius to Kelvin Conversion
Converting from Celsius to Kelvin is a basic yet crucial step when working with gas laws. As described earlier, this conversion is necessary because Kelvin is the preferred unit of measure for scientific purposes, where calculations require absolute temperatures.Let’s reiterate the process:
- Add 273.15 to the Celsius temperature to convert it to Kelvin.
- For instance, \( 19^{\circ} \)C becomes \( 292.15 \) K and \( 58^{\circ} \)C is \( 331.15 \) K.
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