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Chapter 19: Problem 67

In a cylinder sealed with a piston, you rapidly compress 3.00 \(\mathrm{L}\) of \(\mathrm{N}_{2}\) gas initially at 1.00 atm pressure and \(0.00^{\circ} \mathrm{C}\) to half its original volume. Assume the \(\mathrm{N}_{2}\) behaves like an ideal gas. (a) Calculate the final temperature and pressure of the gas. (b) If you now cool the gas back to \(0.00^{\circ} \mathrm{C}\) without changing the pressure, what is its final volume?

Short Answer

Expert verified
(a) Final pressure: 2.00 atm, Final temperature: 273.15 K. (b) Final volume: 0.75 L.

Step by step solution

01

Initial Setup and Known Values

We know that the initial volume of the gas, \(V_1\), is 3.00 L. The initial pressure, \(P_1\), is 1.00 atm and the temperature, \(T_1\), is 0.00掳C, which is 273.15 K when converted to Kelvin. We are given that the volume is compressed to half, making the final volume, \(V_2\), equal to 1.50 L.
02

Applying the Ideal Gas Law

The ideal gas law, \(PV = nRT\), can be used in its combined form for initial and final states, expressed as \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). We need to find both the final temperature, \(T_2\), and the final pressure, \(P_2\).
03

Calculating the Final Pressure (\(P_2\))

Using Boyle's Law (since temperature changes haven't been accounted yet), where \(P_1V_1 = P_2V_2\), solve for \(P_2\): \[P_2 = \frac{P_1V_1}{V_2} = \frac{1.00\, \text{atm} \times 3.00\, \text{L}}{1.50\, \text{L}} = 2.00\, \text{atm}\]. Thus, the final pressure \(P_2 = 2.00\, \text{atm}\).
04

Calculating the Final Temperature (\(T_2\))

Using the combined gas law \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\), solve for \(T_2\): \[T_2 = \frac{P_2V_2T_1}{P_1V_1} = \frac{2.00\, \text{atm} \times 1.50\, \text{L} \times 273.15\, \text{K}}{1.00\, \text{atm} \times 3.00\, \text{L}} = 273.15\, \text{K}\]. Therefore, the final temperature is 273.15 K.
05

Determining Final Volume After Cooling to 0掳C

Cooling the gas at constant pressure back to \(0.00^{\circ}\text{C}\) (273.15 K), apply Charles's Law, where \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), to find the new volume, \(V\): \[V = \frac{T_2 \times V_2}{T_1} = \frac{273.15\, \text{K} \times 1.50\, \text{L}}{546.30\, \text{K}} = 0.75\, \text{L}\]. Thus, the gas volume is reduced to 0.75 L.

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Key Concepts

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

Boyle's Law and its Relation to Gas Pressure and Volume
Boyle's Law is an important concept in understanding the behavior of gases. It describes how the pressure of a gas tends to increase as the volume decreases, provided the temperature remains constant. The law is mathematically expressed as \( P_1 V_1 = P_2 V_2 \). Here, \( P \) represents the pressure and \( V \) the volume of the gas. This direct relationship tells us that if the volume of a gas is reduced, its pressure rises, assuming no change in temperature. For example, when compressing the nitrogen gas in a sealed cylinder, its volume decreases while its pressure increases. By applying Boyle鈥檚 Law, we determined that when the volume was halved from 3.00 L to 1.50 L, the initial pressure of 1.00 atm rose to 2.00 atm.
Charles's Law and Volume Change with Temperature
Charles's Law provides insight into how gases expand and contract with temperature change. It shows that the volume of a gas is directly proportional to its temperature measured in Kelvin, as long as the pressure remains constant. The formula \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) describes this relationship, where \( V \) is the volume and \( T \) is the temperature in Kelvin. When the compressed nitrogen gas was cooled back to 0掳C, Charles's Law was applied to find the new volume. Even though the pressure was kept constant, cooling the gas caused it to contract from 1.50 L to 0.75 L, hence highlighting how temperature changes can impact gas volume considerably.
Understanding Nitrogen Gas in Ideal Gas Law Applications
Nitrogen gas often serves as a common example in demonstrations of the Ideal Gas Law because it typically behaves like an ideal gas under many conditions. The Ideal Gas Law equation \( PV = nRT \) relates pressure (\( P \)), volume (\( V \)), the number of moles (\( n \)), the ideal gas constant (\( R \)), and temperature (\( T \)) to describe the state of a gas. In our scenario, nitrogen gas in a piston-cylinder setup allowed us to explore various thermodynamic changes. It illustrates how gases, similarly, follow predictable patterns under varying conditions of pressure, volume, and temperature. Even though real gases may not perfectly adhere to ideal behavior, such approximations enable practical applications in evaluating, predicting, or calculating gas behavior in diverse fields.

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Most popular questions from this chapter

Five moles of an ideal monatomic gas with an initial temperature of \(127^{\circ} \mathrm{C}\) expand and, in the process, absorb 1200 \(\mathrm{J}\) of heat and do 2100 \(\mathrm{J}\) of work. What is the final temperature of the gas?

A player bounces a basketball on the floor, connpressing it to 80.0\(\%\) of its original volume. The air (assume it is essentially N_{2} gas) inside the ball is originally at a temperature of \(20.0^{\circ} \mathrm{C}\) and a pressure of 2.00 \(\mathrm{atm}\) . The ball's diameter is \(23.9 \mathrm{cm} .\) (a) What temperature does the air in the ball reach at its maximum compression? (b) By how much does the internal energy of the air change between the ball's original state and its maximum compression?

An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

On a warm summer day, a large mass of air (atmospheric pressure \(1.01 \times 10^{5} \mathrm{Pa}\) ) is heated by the ground to a temperature of \(26.0^{\circ} \mathrm{C}\) and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why? Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only \(0.850 \times 10^{5} \mathrm{Pa}\) . Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cool- ing for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of altitude, is called the dry adiabatic lapse rate.)

Oscillations of a Piston. A vertical cylinder of radins \(r\) contains a quantity of ideal gas and is fitted with a piston with mass \(m\) that is free to move (Fig. 19.34\()\) . The piston and the walls of the cylinder are frictionless and the entire cylinder is placed in aconstant-temperature bath. The outside air pressure is \(p_{0}\) . In equilibrium, the piston sits at a height \(h\) above the bottom of the cylinder. (a) Find the absolute pressure of the gas trapped below the piston when in equilibrium. (b) The piston is pulled up by a small distance and released. Find the net force acting on the piston when its base is a distance \(h+y\) above the bottom of the cylinder, where \(y\) is much less than \(h\) (c) After the piston is displaced from equilibrium and released, it oscillates up and down. Find the frequency of nthese small oscillations. If the displacement is not small, are the oscillations simple harmonic? How can you tell?
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