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Ch 18: A Macroscopic Description of Matter
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 18, Problema 72

An inflated bicycle inner tube is 2.2 cm in diameter and 200 cm in circumference. A small leak causes the gauge pressure to decrease from 110 psi to 80 psi on a day when the temperature is 20°C. What mass of air is lost? Assume the air is pure nitrogen.

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Determine the initial and final pressures in absolute terms by adding atmospheric pressure (approximately 14.7 psi) to the gauge pressures. Convert these pressures to Pascals (Pa) using the conversion factor: 1 psi = 6894.76 Pa.
Use the ideal gas law, \( PV = nRT \), to relate the initial and final states of the gas. Here, \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (8.314 J/(mol·K)), and \( T \) is temperature in Kelvin. Convert the temperature from Celsius to Kelvin using \( T(K) = T(°C) + 273.15 \).
Calculate the volume of the inner tube. Treat the tube as a cylindrical shell. The volume \( V \) can be approximated as \( V = \pi r^2 h \), where \( r \) is the radius (half the diameter) and \( h \) is the circumference. Convert the dimensions to meters before calculating.
Using the ideal gas law, calculate the initial number of moles \( n_1 \) and the final number of moles \( n_2 \) of nitrogen in the inner tube. Rearrange the ideal gas law to solve for \( n \): \( n = \frac{PV}{RT} \). Use the initial and final pressures, the calculated volume, and the temperature in Kelvin.
Find the mass of air lost by calculating the difference in moles \( \Delta n = n_1 - n_2 \) and multiplying by the molar mass of nitrogen (28.02 g/mol). Convert the result to kilograms by dividing by 1000.

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Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This law is essential for understanding how changes in pressure and temperature affect the volume and amount of gas present. In this scenario, it helps determine how the decrease in pressure due to the leak correlates with the loss of air mass.
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Ideal Gases and the Ideal Gas Law

Gauge Pressure

Gauge pressure is the pressure of a system above atmospheric pressure. It is measured relative to the ambient atmospheric pressure and is crucial for understanding the pressure changes in the inner tube. In this question, the initial and final gauge pressures indicate how much air has escaped due to the leak.
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Pressure Gauges: Barometer

Density of Nitrogen

The density of nitrogen at standard conditions is approximately 1.25 kg/m³. Knowing the density is vital for calculating the mass of the air lost when the pressure changes. By using the Ideal Gas Law and the change in volume due to the pressure drop, one can find the volume of nitrogen lost and subsequently its mass.
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Calculating Radius of Nitrogen
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The closed cylinder of FIGURE CP18.74 has a tight-fitting but frictionless piston of mass M. The piston is in equilibrium when the left chamber has pressure p₀ and length L₀ while the spring on the right is compressed by ΔL. Suppose the piston is moved a small distance x to the right. Find an expression for the net force (Fₓ)net on the piston. Assume all motions are slow enough for the gas to remain at the same temperature as its surroundings.

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The cylinder in FIGURE CP18.73 has a moveable piston attached to a spring. The cylinder's cross-section area is 10 cm2, it contains 0.0040 mol of gas, and the spring constant is 1500 N/m. At 20°C the spring is neither compressed nor stretched. How far is the spring compressed if the gas temperature is raised to 100°C?

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Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20°C undergo an isobaric expansion until the volume has tripled. What is the gas temperature after the expansion (in °C)? The gas pressure is then decreased at constant volume until the original temperature is reached.

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In Problems 67,68,69,67, 68, 69, and 7070 you are given the equation(s) used to solve a problem. For each of these, you are to write a realistic problem for which this is the correct equation(s).

(T2+273) K=200 kPa500 kPa×1×(400+273) K(T_2 + 273) \(\text{ K}\) = \(\frac{200 \text{ kPa}\)}{500 \(\text{ kPa}\)} \(\times\) 1 \(\times\) (400 + 273) \(\text{ K}\)

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In Problems 67,68,69,67, 68, 69, and 7070 you are given the equation(s) used to solve a problem. For each of these, you are to write a realistic problem for which this is the correct equation(s).

p2=300 cm3100 cm3×1×2 atmp_2 = \(\frac{300 \text{ cm}\)^3}{100 \(\text{ cm}\)^3} \(\times\) 1 \(\times\) 2 \(\text{ atm}\)

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