/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 The gauge pressure in a car tire... [FREE SOLUTION] | 91Ó°ÊÓ

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The gauge pressure in a car tire is \(305 \mathrm{kPa}\) when its temperature is \(15^{\circ} \mathrm{C}\). After running at high speed, the tire has heated up and its gauge pressure is \(360 \mathrm{kPa}\). What is then the temperature of the gas in the tire? Assume atmospheric pressure to be \(101 \mathrm{kPa}\). Being careful to use only absolute temperature and absolute pressures: $$ \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \quad \text { or } \quad T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)\left(\frac{V_{2}}{V_{1}}\right) $$ $$ \text { with } \quad P_{1}=305 \mathrm{kPa}+101 \mathrm{kPa}=406 \mathrm{kPa} \quad \text { and } \quad P_{2}=360 \mathrm{kPa}+101 \mathrm{kPa}=461 \mathrm{kPa} $$ $$ \begin{array}{l} \text { Then }\\\ T_{2}=(273+15)\left(\frac{461}{406}\right)(1.00)=327 \mathrm{~K} \end{array} $$ The final temperature of the tire is \(327-273=54{ }^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The temperature of the gas in the tire after heating is \(54^{\circ} \mathrm{C}\).

Step by step solution

01

Convert Gauge Pressure to Absolute Pressure

Gauge pressure is given as \(305 \mathrm{kPa}\), but we need the absolute pressure. To convert, add the atmospheric pressure which is \(101 \mathrm{kPa}\). Therefore, \(P_1 = 305 + 101 = 406 \mathrm{kPa}\). Similarly, the final gauge pressure is \(360 \mathrm{kPa}\), so \(P_2 = 360 + 101 = 461 \mathrm{kPa}\).
02

Convert Initial Temperature to Kelvin

Initial temperature is given in Celsius as \(15^{\circ} \mathrm{C}\). Convert this to Kelvin using the formula \(T(K) = T(^{\circ}C) + 273\). So, \(T_1 = 15 + 273 = 288 \mathrm{K}\).
03

Determine Relationship Between Initial and Final States

We assume the volume \(V\) remains constant \((V_1 = V_2)\). Using the ideal gas law in the form \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\), we derive \(T_2 = T_1 \left( \frac{P_2}{P_1} \right)\).
04

Calculate Final Temperature in Kelvin

Plug the known values into the formula: \(T_2 = 288 \left( \frac{461}{406} \right) \) to find \(T_2 = 327 \mathrm{K}\).
05

Convert Final Temperature to Celsius

The final temperature in Kelvin is \(327 \mathrm{K}\). To convert this back to Celsius, subtract 273: \(T_2 = 327 - 273 = 54^{\circ} \mathrm{C}\).

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Key Concepts

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

Gauge Pressure
In our daily lives, we often encounter things like car tire pressure which is measured using gauge pressure. Gauge pressure is the pressure measurement that excludes atmospheric pressure and often shows us how much above or below the atmospheric pressure a particular system is. When we fill up a car tire, the gauge tells us how much over the local atmospheric pressure the tire is filled. So, if a gauge reads 305 kPa, this means there is 305 kPa more than the surrounding atmospheric pressure within the tire.

Understanding gauge pressure is crucial because most pressure gauges don't account for atmospheric pressure, meaning every pressure reading assumes the atmospheric pressure is zero. This is helpful for applications like filling tires but when calculating changes in temperature or other conditions using the ideal gas law, we must use absolute pressure.
Absolute Temperature
Absolute temperature is a vital concept in the field of thermodynamics and relates to the absolute energy a particle system holds. It is measured on the Kelvin scale, which starts from absolute zero. Unlike Celsius or Fahrenheit, there are no negative numbers in Kelvin, which makes it very useful for scientific calculations. Absolute zero is theoretically the point where molecular movement stops—the lowest possible energy state.
To convert from Celsius to Kelvin, you simply add 273. For instance, 15°C becomes 288K (15 + 273 = 288). Understanding absolute temperature is important because it ensures that calculations involving gas laws, like the Ideal Gas Law, hold true regardless of the units of measurement originally provided.
Pressure Conversion
Understanding how to convert pressures from gauge to absolute is a key part of many physics and engineering calculations. Since pressure gauges typically measure the energy or pressure difference compared to atmospheric conditions, adding atmospheric pressure to gauge pressure results in absolute pressure. For most calculations using the Ideal Gas Law, absolute pressure is necessary. This is because the law mathematically connects pressure, volume, and absolute temperature in a way that needs the full force or total pressure exerted by the gas, which includes Earth's atmospheric pressure.

For example, converting a gauge pressure of 305 kPa involves adding the atmospheric pressure, which is usually about 101 kPa. This gives an absolute pressure of 406 kPa (305 + 101). Similarly, for the final pressure calculation detailed in our exercise, we convert with the same method: 360 kPa gauge pressure becomes 461 kPa absolute pressure.
Temperature Conversion
The conversion of temperature from Celsius, common in everyday settings, to Kelvin, necessary in scientific calculations, is a straightforward but critical process. Since the Ideal Gas Law requires temperatures to be in Kelvin, knowing how to make this conversion is crucial.
  • To convert from Celsius to Kelvin, add 273 to the Celsius temperature.
  • If you have a Kelvin temperature and need to convert it to Celsius, you subtract 273.
For example, an initial temperature of 15°C is converted to 288K by adding 273 (15 + 273 = 288). Similarly, once calculations give you a temperature like 327K, converting back to Celsius involves subtracting 273 to result in 54°C (327 - 273 = 54). Always remember, the constant of 273 smooths out the scale difference between Kelvin and Celsius, ensuring accuracy.

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Most popular questions from this chapter

Compute the volume of \(8.0\) g of helium \((M=4.0 \mathrm{~kg} / \mathrm{kmol})\) at 15 \({ }^{\circ} \mathrm{C}\) and \(480 \mathrm{mmHg}\). Use \(P V=(m / M) R T\) to obtain $$ V=\frac{m R T}{M P}=\frac{(0.0080 \mathrm{~kg})(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(288 \mathrm{~K})}{(4.0 \mathrm{~kg} / \mathrm{kmol})\left[(480 / 760)\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\right]}=0.075 \mathrm{~m}^{3}=75 \text { liters } $$

A fish emits a \(2.0-\mathrm{mm}^{3}\) bubble at a depth of \(15 \mathrm{~m}\) in a lake. Find the volume of the bubble as it reaches the surface. Assume its omnoraturo dooc not chonore The absolute pressure in the bubble at a depth \(h\) is $$ P=\rho g h+\text { Atmospheric pressure } $$ where \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\) and atmospheric pressure is about \(100 \mathrm{kPa}\). At \(15 \mathrm{~m}\), $$ P_{1}=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{3}\right)(15 \mathrm{~m})+100 \mathrm{kPa}=247 \mathrm{kPa} $$ and at the surface, \(P_{2}=100 \mathrm{kPa}\). Following the usual procedure, $$ V_{2}=V_{1}\left(\frac{P_{1}}{P_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right)=\left(2.0 \mathrm{~mm}^{3}\right)\left(\frac{247}{100}\right)(1.0)=4.9 \mathrm{~mm}^{3} $$

An ideal gas is in a chamber at a pressure of \(2.00 \mathrm{MPa}\) and has a volume of \(20.0\) liters when at a temperature of \(298.15 \mathrm{~K}\). Determine the number of moles of gas in the chamber.

An ideal gas has a volume of exactly 1 liter at \(1.00 \mathrm{~atm}\) and \(-20{ }^{\circ} \mathrm{C}\). To how many atmospheres of pressure must it be subjected in order to be compressed to \(0.500\) liter when the temperature is 40 \({ }^{\circ} \mathrm{C}\) ? $$ \begin{array}{c} \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \quad \text { or } \quad P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right) \\\ \text { from which } \quad P_{2}=(1.00 \mathrm{~atm})\left(\frac{1.00 \mathrm{~L}}{0.500 \mathrm{~L}}\right)\left(\frac{273 \mathrm{~K}+40 \mathrm{~K}}{273 \mathrm{~K}-20 \mathrm{~K}}\right)=2.47 \mathrm{~atm} \end{array} $$

A sealed tank having a volume of \(25.0 \times 10^{-3} \mathrm{~m}^{3}\) contains \(0.56 \mathrm{~kg}\) of nitrogen \(\left(\mathrm{N}_{2}\right)\). How many kilomoles of gas are in the tank? [Hint: The atomic mass of the molecule is 28.]

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