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Chapter 2: Problem 34

(a) What is the gauge pressure in a \(25.0^{\circ} \mathrm{C}\) car tire containing 3.60 mol of gas in a 30.0 -L volume? (b) What will its gauge pressure be if you add 1.00 L of gas originally at atmospheric pressure and \(25.0^{\circ} \mathrm{C}\) ? Assume the temperature remains at \(25.0^{\circ} \mathrm{C}\) and the volume remains constant.

Short Answer

Expert verified
The initial gauge pressure in the car tire is \(1.47 脳 10^5\ \mathrm{Pa}\). After adding 1.00 L of gas at atmospheric pressure, the new gauge pressure is \(1.52 脳 10^5\ \mathrm{Pa}\).

Step by step solution

01

Part (a): Calculate Initial Gauge Pressure using the Ideal Gas Law

The ideal gas law is: \(PV = nRT\) Where: - P = pressure (Pa) - V = volume (L) - n = number of moles of gas (mol) - R = ideal gas constant (8.314 J/mol路K) - T = temperature (K) First, we need to convert the temperature from Celsius to Kelvin: \(T_K = 25.0 + 273.15 = 298.15\ \mathrm{K}\) Now, we can rearrange the ideal gas law formula to find the pressure: \(P = \frac{nRT}{V}\) We are given: - n = 3.60 mol - R = 8.314 J/mol路K - T = 298.15 K - V = 30.0 L Now substitute the values and find the pressure: \(P = \frac{(3.60\ \mathrm{mol})(8.314\ \mathrm{J/mol路K})(298.15\ \mathrm{K})}{(30.0\ \mathrm{L})}\) Note: Convert volume from L to m鲁: \(V_{m^3} = 30.0\ \mathrm{L} 脳 \frac{1\ \mathrm{m^3}}{1000\ \mathrm{L}} = 0.030\ \mathrm{m^3}\) \(P = \frac{(3.60\ \mathrm{mol})(8.314\ \mathrm{J/mol路K})(298.15\ \mathrm{K})}{(0.030\ \mathrm{m^3})} = 2.48 脳 10^5\ \mathrm{Pa}\) The gauge pressure is the pressure relative to atmospheric pressure, which is \(1.013 脳 10^5\ \mathrm{Pa}\). Thus, the initial gauge pressure is: \(P_{gauge} = P - P_{atm} = (2.48 - 1.013) 脳 10^5\ \mathrm{Pa} = 1.47 脳 10^5\ \mathrm{Pa}\)
02

Part (b): Find New Gauge Pressure after Adding Gas

To find the new gauge pressure, first find the number of moles in the added 1.00 L of gas at atmospheric pressure. Using the ideal gas law, we can find the new total number of moles in the tire and the new pressure. The added gas has a volume of 1.00 L, and it's at atmospheric pressure, so we can use the ideal gas law to find the number of moles in the added gas: \(n = \frac{PV}{RT}\) We are given: - P = atmospheric pressure (\(1.013 脳 10^5\ \mathrm{Pa}\)) - V = 1.00 L - R = 8.314 J/mol路K - T = 298.15 K \(n_{added} = \frac{(1.013 脳 10^5\ \mathrm{Pa})(0.001\ \mathrm{m^3})}{(8.314\ \mathrm{J/mol路K})(298.15\ \mathrm{K})} = 0.0408\ \mathrm{mol}\) Now find the new total number of moles in the tire: \(n_{total} = n + n_{added} = 3.60\ \mathrm{mol} + 0.0408\ \mathrm{mol} = 3.6408\ \mathrm{mol}\) Now, use the ideal gas law again to find the new pressure: \(P_{new} = \frac{n_{total}RT}{V}\) \(P_{new} = \frac{(3.6408\ \mathrm{mol})(8.314\ \mathrm{J/mol路K})(298.15\ \mathrm{K})}{(0.030\ \mathrm{m^3})} = 2.53 脳 10^5\ \mathrm{Pa}\) Finally, find the new gauge pressure: \(P_{new\ gauge} = P_{new} - P_{atm} = (2.53 - 1.013) 脳 10^5\ \mathrm{Pa} = 1.52 脳 10^5\ \mathrm{Pa}\)

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Key Concepts

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

Gauge Pressure
When we talk about the pressure of a gas, like in a car tire, it's important to differentiate between absolute pressure and gauge pressure. Absolute pressure is the total pressure within a system, including the external atmospheric pressure. In contrast, gauge pressure is the pressure difference between the absolute pressure within the system and the atmospheric pressure outside it. Think of gauge pressure as the extra pressure a gas exerts, not counting what's already exerted by the atmosphere.

For our tire, the gauge pressure is found by subtracting the atmospheric pressure from the calculated tire pressure. It's what tire pressure gauges measure and what we're interested in when pumping our tires. If the gauge reads zero, it doesn't mean there's no pressure inside, but rather that the pressure is equal to the atmospheric pressure. This understanding is crucial to ensure safety and optimal performance in applications like tire inflation and various industrial processes.
Gas Laws
Understanding the behavior of gases involves several fundamental principles, collectively known as gas laws. These laws describe how various properties of a gas鈥攊ts pressure (P), volume (V), temperature (T), and number of moles (n)鈥攁re interrelated. One of the most comprehensive equations that consolidates these properties into a single expression is the Ideal Gas Law: \( PV = nRT \).

This equation tells us that the pressure and volume of a gas are directly proportional to the number of moles and the temperature, as long as the gas behaves ideally鈥攁n approximation that applies well under typical conditions, like at room temperature and atmospheric pressure. Moreover, the Ideal Gas Law is versatile; it aids in calculations of pressure changes during temperature fluctuations or volume adjustments, features that are critical in many scientific and engineering applications, including our tire pressure example.
Molar Volume
Molar Volume is a term used to describe the volume that one mole of a substance occupies at a given temperature and pressure. In the context of an ideal gas, molar volume provides a handy way to understand the space requirements of a gas or to estimate the amount of gas needed to fill a particular volume.

Under standard temperature and pressure (STP, which is 273.15 K and 1 atm), one mole of an ideal gas occupies approximately 22.4 liters. However, this value changes with temperature and pressure variances. The relationship between the molar volume and the Ideal Gas Law is intrinsic. When working on problems, like adding gas to a car tire, thinking in terms of molar volume can simplify the approach鈥攁s when we calculate the number of moles in a known volume of gas at a specific temperature and pressure.
Temperature Conversion
In the study of gas laws, temperature plays a vital role as it affects how molecules within the gas move and thus the pressure they exhibit. It's crucial to use the right temperature scale; for the Ideal Gas Law, we always convert temperature into Kelvin (K). The Kelvin scale is an absolute temperature scale, meaning it starts at absolute zero, the lowest possible temperature where molecular motion ceases.

To convert Celsius to Kelvin, which is a common requirement because most ambient temperature measurements are in Celsius, we use the simple relationship: \( T_K = T_C + 273.15 \), where \( T_C \) is the Celsius temperature and \( T_K \) the equivalent in Kelvin. This conversion ensures accuracy in gas law calculations, as seen in the car tire pressure problem, where we initially corrected the given Celsius temperature to Kelvin to use the Ideal Gas Law properly.

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Most popular questions from this chapter

One cylinder contains helium gas and another contains krypton gas at the same temperature. Mark each of these statements true, false, or impossible to determine from the given information. (a) The ms speeds of atoms in the two gases are the same. (b) The average kinetic energies of atoms in the two gases are the same. (c) The internal energies of 1 mole of gas in each cylinder are the same. (d) The pressures in the two cylinders are the same.

Experimentally it appears that many polyatomic molecules' vibrational degrees of freedom can contribute to some extent to their energy at room temperature. Would you expect that fact to increase or decrease their heat capacity from the value \(R\) ? Explain.

(a) Find the density in SI units of air at a pressure of 1.00 atm and a temperature of \(20^{\circ} \mathrm{C}\), assuming that air is \(78 \% \mathrm{N}_{2}, 21 \% \mathrm{O}_{2},\) and \(1 \% \mathrm{Ar},\) (b) Find the density of the atmosphere on Venus, assuming that it's \(96 \% \mathrm{CO}_{2}\) and \(4 \% \mathrm{N}_{2},\) with a temperature of \(737 \mathrm{K}\) and a pressure of 92.0 atm.

A person hits a tennis ball with a mass of \(0.058 \mathrm{kg}\) against a wall. The average component of the ball's velocity perpendicular to the wall is \(11 \mathrm{m} / \mathrm{s}\), and the ball hits the wall every \(2.1 \mathrm{s}\) on average, rebounding with the opposite perpendicular velocity component. (a) What is the average force exerted on the wall? (b) If the part of the wall the person hits has an area of \(3.0 \mathrm{m}^{2}\), what is the average pressure on that area?

The gauge pressure in your car tires is 18. The gase \(2.50 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) at a temperature of \(35.0^{\circ} \mathrm{C}\) when you drive it onto a ship in Los Angeles to be sent to Alaska. What is their gauge pressure on a night in Alaska when their temperature has dropped to \(-40.0^{\circ} \mathrm{C}\) ? Assume the tires have not gained or lost any air. 18. The gauge pressure in your car tires is 18. The gase \(2.50 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) at a temperature of \(35.0^{\circ} \mathrm{C}\) when you drive it onto a ship in Los Angeles to be sent to Alaska. What is their gauge pressure on a night in Alaska when their temperature has dropped to \(-40.0^{\circ} \mathrm{C}\) ? Assume the tires have not gained or lost any air.
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