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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 hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is \(500.0 \mathrm{~m}^{3}\) and the surrounding air is at \(15.0^{\circ} \mathrm{C}\) what must the temperature of the air in the balloon be for it to lift a total load of \(290 \mathrm{~kg}\) (in addition to the mass of the hot air)? The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is \(1.23 \mathrm{~kg} / \mathrm{m}^{3}\).

Helium gas is in a cylinder that has rigid walls. If the pressure of the gas is 2.00 atm, then the root-mean-square speed of the helium atoms is \(v_{\mathrm{rms}}=176 \mathrm{~m} / \mathrm{s} .\) By how much (in atmospheres) must the pressure be increased to increase the \(v_{\mathrm{rms}}\) of the He atoms by \(100 \mathrm{~m} / \mathrm{s} ?\) Ignore any change in the volume of the cylinder.

Oxygen \(\left(\mathrm{O}_{2}\right)\) has a molar mass of \(32.0 \mathrm{~g} / \mathrm{mol} .\) What is (a) the average translational kinetic energy of an oxygen molecule at a temperature of \(300 \mathrm{~K} ;\) (b) the average value of the square of its speed; (c) the rootmean-square speed; (d) the momentum of an oxygen molecule traveling at this speed? (e) Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel \(0.10 \mathrm{~m}\) on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) (f) What is the average force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm? (h) Compute the number of oxygen molecules that are contained in a vessel of this size at \(300 \mathrm{~K}\) and atmospheric pressure. (i) Your answer for part (h) should be three times as large as the answer for part \((\mathrm{g})\). Where does this discrepancy arise?

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 \(15.0^{\circ} \mathrm{C} ?\)

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains \(0.110 \mathrm{~m}^{3}\) of air at a pressure of 0.355 atm. The piston is slowly pulled out until the volume of the gas is increased to \(0.390 \mathrm{~m}^{3}\). If the temperature remains constant, what is the final value of the pressure?
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