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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?

Short Answer

Expert verified
The new pressure in the bicycle tire is approximately \(79.52\) psi.

Step by step solution

01

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: \(T_2 = 58 + 273.15 = 331.15 K\)
02

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 \(V_2 = V_1 + 0.04 V_1 = 1.04V_1\).
03

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 \(P_2\): \(P_2 = \frac{P_1V_1T_2}{T_1V_2}\)
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
For example, if you have a temperature of \(19^{\circ}\)C, you add 273.15 to get \(292.15\) K. Similarly, \(58^{\circ}\)C becomes \(331.15\) K. This straightforward conversion is crucial because the Ideal Gas Law requires temperature in Kelvin to accurately describe the behavior of gases.
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.
Understanding pressure calculation through this formula helps you visualize how gas laws underpin everyday phenomena like a bike tire heating up in the sun.
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 \( V_2 = 1.04 \times V_1 \).
This formula comes from understanding that a 4% increase means adding 4% of the original volume to itself. It reveals how volume expansion due to temperature can affect pressure in enclosed structures like tires, which get hotter on sunny days, causing them to expand.
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.
Using Kelvin ensures consistent results with gas laws because it provides a true zero—the point at which kinetic energy of particles is minimal—which is crucial for precise scientific calculations. Regular practice of this conversion helps solidify understanding of temperature's role in pressure and volume changes in gases.

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Most popular questions from this chapter

The oxides of Group \(2 \mathrm{A}\) metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$\mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s)$$ A \(2.85\) \(-\mathrm{g}\) sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a 3.00 -L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of \(740 .\) torr at \(20 .^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is \(390 .\) torr at \(20 .^{\circ} \mathrm{C}\). What is the mass percent of \(MgO\) in the mixture? Assume that only the \(MgO\) reacts with \(\mathrm{CO}_{2}\)

From the values in Table \(8-3\) for the van der Waals constant \(a\) for the gases \(\mathrm{H}_{2}, \mathrm{CO}_{2}, \mathrm{N}_{2},\) and \(\mathrm{CH}_{4},\) predict which of these gas molecules show the strongest intermolecular attractions.

Assume that \(4.19 \times 10^{6} \mathrm{kJ}\) of energy is needed to heat a home. If this energy is derived from the combustion of methane \(\left(\mathrm{CH}_{4}\right),\) what volume of methane, measured at 1.00 atm and \(0^{\circ} \mathrm{C},\) must be burned? \(\left(\Delta H_{\text {combustion }}^{\circ} \text { for } \mathrm{CH}_{4}=-891 \mathrm{kJ} / \mathrm{mol}\right)\).

An important process for the production of acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is given by the following equation: $$2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ A \(150 .\)-\(\mathrm{L}\)reactor is charged to the following partial pressures at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned}P_{\mathrm{C}, \mathrm{H}_{6}} &=0.500 \mathrm{MPa} \\\P_{\mathrm{NH}_{3}} &=0.800 \mathrm{MPa} \\\P_{\mathrm{O}_{2}} &=1.500 \mathrm{MPa}\end{aligned}$$ What mass of acrylonitrile can be produced from this mixture \(\left(\mathrm{MPa}=10^{6} \mathrm{Pa}\right) ?\)

A piece of solid carbon dioxide, with a mass of \(7.8 \mathrm{g},\) is placed in a \(4.0\)-\(\mathrm{L}\) otherwise empty container at \(27^{\circ} \mathrm{C}\). What is the pressure in the container after all the carbon dioxide vaporizes? If \(7.8 \mathrm{g},\) solid carbon dioxide were placed in the same container but it already contained air at \(740\) torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide vaporizes?

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