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Chapter 18: Problem 7

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains \(499 \mathrm{~cm}^{3}\) of air at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{~Pa}\right)\) and a temperature of \(27.0^{\circ} \mathrm{C}\). At the end of the stroke, the air has been compressed to a volume of \(46.2 \mathrm{~cm}^{3}\) and the gauge pressure has increased to \(2.72 \times 10^{6} \mathrm{~Pa}\). Compute the final temperature.

Short Answer

Expert verified
The final temperature of the air in the cylinder after compression is approximately \(839.76 K\).

Step by step solution

01

Convert units and temperatures

First, convert the initial volume (499 cm鲁) to m鲁 using the conversion factor \(1 m^3 = 1 \times 10^6 cm^3\). The result is \(V1 = 499 \times 10^{-6} m^3\). The final volume (46.2 cm鲁) is converted in the same way, \(V2 = 46.2 \times 10^{-6} m^3\). The initial and final pressures, \(P1 = 1.01 \times 10^5 Pa\) and \(P2 = 2.72 \times 10^6 Pa +1.01 \times 10^5 Pa\), are already in Pa. The gauge pressure does not include atmospheric pressure, so it needs to be added to the final pressure. Convert the initial temperature (27.0掳C) to K using the conversion \(T[K] = T[C] + 273.15\), resulting in \(T1 = 27.0 + 273.15 = 300.15 K\). Solve for final temperature T2, which is unknown.
02

Apply the ideal gas law

Now, we can use the ideal gas law to relate the initial and final conditions. This law is written as \(P1 * V1 / T1 = P2 * V2 / T2\), where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures. Solving this equation for the final temperature gives \(T2 = P2 * V2 * T1 / (P1 * V1)\).
03

Compute the final temperature

Inserting the known values into the equation from step 2 gives \(T2 = 2.82 \times 10^6 Pa * 46.2 \times 10^{-6} m^3 * 300.15 K / (1.01 \times 10^5 Pa * 499 \times 10^{-6} m^3) = 839.76 K\). Therefore, the final temperature of the air in the cylinder after compression is approximately 839.76 K.

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Key Concepts

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

Temperature Conversion
Converting temperature from Celsius to Kelvin is a crucial step when dealing with gas laws. In scientific calculations involving gases, temperature must be in Kelvin. This is because the Kelvin scale starts at absolute zero, the point at which all molecular motion ceases, providing a natural baseline for calculations. To convert Celsius to Kelvin, simply add 273.15 to the Celsius value. For example, an initial temperature of \(27^\circ C\) converts to:
  • \(27 + 273.15 = 300.15\) K
Kelvin is the standard unit for temperature in the scientific community, particularly useful when applying the ideal gas law, which requires uniformity in units across pressure, volume, and temperature.
Volume Conversion
Understanding volume conversion is essential when working with gas laws, such as the ideal gas law. Volumes are often given in cubic centimeters (cm鲁) but need to be in cubic meters (m鲁) for compatibility with other units like pressure in Pascals. The conversion factor is simple:
  • \(1 \text{ m}^3 = 1 \times 10^6 \text{ cm}^3\)
To convert \(499 \text{ cm}^3\) to cubic meters, multiply by \(10^{-6}\), getting \(499 \times 10^{-6} \text{ m}^3 = 0.000499 \text{ m}^3\). Similarly, \(46.2 \text{ cm}^3\) converts to \(46.2 \times 10^{-6} \text{ m}^3 = 0.0000462 \text{ m}^3\). Accurately converting volumes ensures consistency when applying the ideal gas law: \(P V = n R T\).
Pressure Calculation
Calculating pressure accurately is fundamental in the context of the ideal gas law. In many problems, you may encounter a given gauge pressure, which is the pressure above atmospheric pressure. To find the absolute pressure, add the atmospheric pressure to the gauge pressure. In this exercise:
  • Gauge Pressure: \(2.72 \times 10^6\) Pa
  • Atmospheric Pressure: \(1.01 \times 10^5\) Pa
By adding these, the total (absolute) pressure at the end of the compression stroke is \(2.82 \times 10^6\) Pa. Remember, the absolute pressure is what the ideal gas law uses to ensure proper calculations. Correct pressure values help in applying the formula \(P1 \cdot V1 / T1 = P2 \cdot V2 / T2\), which relates the initial and final states of the gas.

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Most popular questions from this chapter

A parcel of air over a campfire feels an upward buoyant force because the heated air is less dense than the surrounding air. By estimating the acceleration of the air immediately above a fire, one can estimate the fire's temperature. The mass of a volume \(V\) of air is \(n M_{\text {air }},\) where \(n\) is the number of moles of air molecules in the volume and \(M_{\text {air }}\) is the molar mass of air. The net upward force on a parcel of air above a fire is roughly given by \(\left(m_{\text {out }}-m_{\text {in }}\right) g,\) where \(m_{\text {out }}\) is the mass of a volume of ambient air and \(m_{\text {in }}\) is the mass of a similar volume of air in the hot zone. (a) Use the ideal-gas law, along with the knowledge that the pressure of the air above the fire is the same as that of the ambient air, to derive an expression for the acceleration \(a\) of an air parcel as a function of \(\left(T_{\text {out }} / T_{\text {in }}\right),\) where \(T_{\text {in }}\) is the absolute temperature of the air above the fire and \(T_{\text {out }}\) is the absolute temperature of the ambient air. (b) Rearrange your formula from part (a) to obtain an expression for \(T_{\text {in }}\) as a function of \(T_{\text {out }}\) and \(a\). (c) Based on your experience with campfires, estimate the acceleration of the air above the fire by comparing in your mind the upward trajectory of sparks with the acceleration of falling objects. Thus you can estimate \(a\) as a multiple of \(g .\) (d) Assuming an ambient temperature of \(15^{\circ} \mathrm{C}\), use your formula and your estimate of \(a\) to estimate the temperature of the fire.

(a) What is the total translational kinetic energy of the air in an empty room that has dimensions \(8.00 \mathrm{~m} \times 12.00 \mathrm{~m} \times 4.00 \mathrm{~m}\) if the air is treated as an ideal gas at 1.00 atm? (b) What is the speed of a \(2000 \mathrm{~kg}\) automobile if its kinetic energy equals the translational kinetic energy calculated in part (a)?

Modern vacuum pumps make it easy to attain pressures of the order of \(10^{-13}\) atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of \(9.00 \times 10^{-14}\) atm and an ordinary temperature of \(300.0 \mathrm{~K}\), how many molecules are present in a volume of \(1.00 \mathrm{~cm}^{3} ?\) (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

A balloon of volume \(750 \mathrm{~m}^{3}\) is to be filled with hydrogen at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{~Pa}\right) .\) (a) If the hydrogen is stored in cylinders with volumes of \(1.90 \mathrm{~m}^{3}\) at a gauge pressure of \(1.20 \times 10^{6} \mathrm{~Pa}\), how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if both the gas in the balloon and the surrounding air are at \(15.0^{\circ} \mathrm{C} ?\) The molar mass of hydrogen \(\left(\mathrm{H}_{2}\right)\) is \(2.02 \mathrm{~g} / \mathrm{mol} .\) The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is \(1.23 \mathrm{~kg} / \mathrm{m}^{3} .\) See Chapter 12 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass \(4.00 \mathrm{~g} / \mathrm{mol}\) ) instead of hydrogen, again at \(15.0^{\circ} \mathrm{C} ?\)

Consider an ideal gas at \(27^{\circ} \mathrm{C}\) and 1.00 atm. To get some idea how close these molecules are to each other, on the average, imagine them to be uniformly spaced, with each molecule at the center of a small cube. (a) What is the length of an edge of each cube if adjacent cubes touch but do not overlap? (b) How does this distance compare with the diameter of a typical molecule? (c) How does their separation compare with the spacing of atoms in solids, which typically are about \(0.3 \mathrm{nm}\) apart?
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