/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 A quantity of \(\mathrm{N}_{2}\)... [FREE SOLUTION] | 91Ó°ÊÓ

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A quantity of \(\mathrm{N}_{2}\) gas originally held at \(531.96 \mathrm{kPa}\) pressure in a 1.00 -L container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5-\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(531.96 \mathrm{kPa}\) and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?

Short Answer

Expert verified
The total pressure in the new container is the sum of the calculated pressures for nitrogen and oxygen.

Step by step solution

01

Understand Initial Conditions for Nitrogen

We begin with a sample of nitrogen gas \(\mathrm{N}_2\), initially held at a pressure of \(P_1 = 531.96\, \mathrm{kPa}\) in a volume \(V_1 = 1.00\, \mathrm{L}\) at a temperature of \(T_1 = 26^{\circ}\, \mathrm{C}\). Convert this temperature to Kelvin by adding 273.15: \(T_{1K} = 26 + 273.15 = 299.15\, \mathrm{K}\).
02

Account for New Conditions for Nitrogen

Transfer the \(\mathrm{N}_2\) gas to a \(V_2 = 12.5\, \mathrm{L}\) container at \(20^{\circ}\, \mathrm{C}\). First, convert the new temperature to Kelvin: \(T_2 = 20 + 273.15 = 293.15\, \mathrm{K}\). Use the Combined Gas Law \(\frac{P_1 \times V_1}{T_{1K}} = \frac{P_2 \times V_2}{T_2}\) to find the new pressure \(P_2\).
03

Calculate New Pressure for Nitrogen

Rearrange the Combined Gas Law to solve for \(P_2\): \(P_2 = \frac{P_1 \times V_1 \times T_2}{T_{1K} \times V_2}\). Substitute the values: \(P_2 = \frac{531.96 \times 1.00 \times 293.15}{299.15 \times 12.5}\). Calculate \(P_2\).
04

Understand Initial Conditions for Oxygen

Now consider \(\mathrm{O}_2\) gas, which is initially at the same pressure \(P_3 = 531.96\, \mathrm{kPa}\), in a volume \(V_3 = 5.00\, \mathrm{L}\), at a temperature \(T_1 = 26^{\circ} \mathrm{C}\) or \(T_{1K} = 299.15\, \mathrm{K}\).
05

Calculate New Pressure for Oxygen

Transfer \(\mathrm{O}_2\) gas to the \(12.5\, \mathrm{L}\) container at \(20^{\circ}\, \mathrm{C}\) (\(T_2 = 293.15\, \mathrm{K}\)). Use the same Combined Gas Law: \(P_4 = \frac{P_3 \times V_3 \times T_2}{T_{1K} \times V_2}\). Substitute the values to find \(P_4\).
06

Calculate the Total Pressure in the Container

With both gases in the \(12.5\, \mathrm{L}\) container, we sum the pressures \(P_2\) (for \(\mathrm{N}_2\)) and \(P_4\) (for \(\mathrm{O}_2\)). The total pressure \(P_{\text{total}} = P_2 + P_4\).

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Key Concepts

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

Gas Laws
Gas laws are fundamental in understanding the behavior of gases under different conditions. In this exercise, we focus on the Combined Gas Law, which is particularly useful when dealing with changing states of a gas, involving pressure (P), volume (V), and temperature (T). The Combined Gas Law is expressed as:\[\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}\]This equation allows us to predict how a gas will respond to changes in temperature, volume, or pressure. When using this law, it's important to remember that temperature must always be in Kelvin. By applying the Combined Gas Law, we can solve problems involving gases moving from one set of conditions to another. It is derived from three other gas laws: Boyle's Law, Charles's Law, and Gay-Lussac's Law, which represent relationships between two variables while holding one constant.
Pressure Calculation
Pressure calculation is a critical part of understanding gas behavior and is often needed in solving gas law problems. Pressure is defined as the force exerted by gas molecules striking the walls of its container, typically measured in units like kPa.In this exercise, we start with an initial pressure of \(531.96\, \mathrm{kPa}\) for both nitrogen and oxygen gases. When transferring the \(\mathrm{N}_2\)gas to a new volume, we calculate the new pressure using:\[P_2 = \frac{P_1 \times V_1 \times T_2}{T_1 \times V_2}\]This relationship helps us understand how force per unit area changes when gases are moved between different volumes and temperatures. For a constant amount of gas, any change in one part of the equation necessitates a change in the others, to maintain equilibrium.
Temperature Conversion
To effectively use the Combined Gas Law, temperatures must always be converted into Kelvin. The Kelvin scale is an absolute temperature scale, where 0 K (-273.15°C) is the point at which molecular motion ceases. In this exercise, the initial temperatures were given in Celsius. For instance, the initial temperature of 26°C is converted to Kelvin by adding 273.15, resulting in 299.15 K. Similarly, the new temperature for both gases after transfer is 20°C, which converts to 293.15 K. The reason Kelvin is used in gas laws is that it avoids negative temperature values, thus simplifying calculations. Accurate temperature conversion is crucial, as it directly impacts the balance of the equation used in finding new gas conditions.
Volume Changes
Volume changes offer insight into how gas spreads out in a container. In this exercise, we deal with the transfer of gases into a shared container and how this impacts the overall pressure. Initially, nitrogen is contained in a 1.00 L volume and is then transferred to a larger 12.5 L container. Similarly, oxygen moves from a 5.00 L volume into the same container. The volume change is significant because the pressure exerted by a gas is inversely related to its volume, according to the principle of Boyle's Law. By transferring the gases, and calculating how each responds to the volume change, we can sum their final pressures to determine the total pressure in the combined volume. Monitoring volume changes is essential not only for pressure calculations but also for understanding how gases interact when mixed.

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Most popular questions from this chapter

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ; \mathrm{in}\) other words, rate is the amount that diffuses over the time it takes to diffuse.)

An ideal gas at a pressure of \(152 \mathrm{kPa}\) is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of \(0.800 \mathrm{~L}\) as shown here. When the stopcock is opened, the gas expands into the empty bulb. If the temperature is held constant during this process and the final pressure is \(92.66 \mathrm{kPa}\), what is the volume of the bulb that was originally filled with gas?

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), liquefies under modest pressure, allowing a large amount to be stored in a container. (a) Calculate the number of moles of propane gas in a 20 -L container at \(709.3 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C} .\) (b) Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is \(0.590 \mathrm{~g} / \mathrm{mL}\). (c) Calculate the ratio of the number of moles of liquid to moles of gas. Discuss this ratio in light of the kinetic-molecular theory of gases.

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of \(126.7 \mathrm{kPa}\) at a temperature of \(-6^{\circ} \mathrm{C} ;(\mathbf{b})\) the absolute temperature of the gas at which \(3.33 \times 10^{-3}\) mol occupies \(478 \mathrm{~mL}\) at \(99.99 \mathrm{kPa} ;(\mathbf{c})\) the pressure, in pascals, if \(0.00245 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138^{\circ} \mathrm{C} ;\) (d) the quantity of gas, in moles, if \(126.5 \mathrm{~L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa} .\)

(a) How high in meters must a column of ethanol be to exert a pressure equal to that of a \(100-\mathrm{mm}\) column of mercury? The density of ethanol is \(0.79 \mathrm{~g} / \mathrm{mL}\), whereas that of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\). (b) What pressure, in atmospheres, is exerted on the body of a diver if she is \(10 \mathrm{~m}\) below the surface of the water when the atmospheric pressure is \(100 \mathrm{kPa}\) ? Assume that the density of the water is \(1.00=1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). The gravitational constant is \(9.81 \mathrm{~m} / \mathrm{s}^{2}\), and \(1 \mathrm{~Pa}=1 \mathrm{~kg} / \mathrm{ms}^{2}\).

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