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A volume of \(190.0 \mathrm{ml}\) of \(\mathrm{N}_{2}\) was collected in a jar over water at some temperature, water level inside and outside the jar standing at the same height. If barometer reads \(740 \mathrm{~mm} \mathrm{Hg}\) and aqueous tension at the temperature of the experiment is \(20 \mathrm{~mm} \mathrm{Hg}\), the volume of the gas at 1 atm pressure and at the same temperature would be (a) \(185.0 \mathrm{ml}\) (b) \(180.0 \mathrm{ml}\) (c) \(195.0 \mathrm{ml}\) (d) \(200 \mathrm{ml}\)

Short Answer

Expert verified
The volume of \(\mathrm{N}_2\) gas at 1 atm pressure is approximately 185.0 ml.

Step by step solution

01

Determine the pressure exerted by the collected gas

Correct for the aqueous tension (water vapor pressure) to find the actual pressure exerted by the nitrogen gas. This can be done by subtracting the aqueous tension from the barometric pressure.
02

Apply Boyle's Law

Use Boyle's Law, which states that at constant temperature, the volume of a fixed amount of gas is inversely proportional to the pressure it exerts, to find the new volume at 1 atm pressure. The formula for Boyle's Law is given by: \[ V_1 \cdot P_1 = V_2 \cdot P_2 \]
03

Calculate the new volume at 1 atm pressure

Rearrange the Boyle's Law equation to solve for the new volume \(V_2\) when the pressure is 1 atm and input the values from Step 1 (pressure exerted by nitrogen only) and the given volume of gas \(V_1\).

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Key Concepts

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

Gas Laws
Understanding gas laws is crucial when working with gases, as they describe how the physical properties of gases—volume, pressure, and temperature—are interrelated. Boyle's Law is one of the fundamental gas laws and it is often applied when discussing the behavior of gases under pressure. Boyle's Law asserts that for a given mass of gas at a constant temperature, the volume of the gas is inversely proportional to its pressure. In simpler terms, if you increase the pressure on a gas, its volume will decrease, provided the temperature remains unchanged, and vice versa. This principle is mathematically expressed as \[ V_1 \cdot P_1 = V_2 \cdot P_2 \], where \(V_1\) and \(P_1\) are the original volume and pressure, respectively, and \(V_2\) and \(P_2\) are the new volume and pressure after a change has been made. Boyle's Law is particularly useful for scientists and engineers in fields such as chemistry, meteorology, and medicine, as it helps predict how gases will behave under different pressures.

The direct application of Boyle's Law often involves a two-step process. First, ascertain the gas's current pressure and volume. Then, if a change in pressure is applied, utilize Boyle's law to determine the new volume of the gas. For students dealing with Boyle's Law problems, it's important to keep in mind the need for pressure units to be consistent; conversion to a common unit such as atmospheres (atm) or millimeters of mercury (mmHg) might be necessary before applying the law.
Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. In other words, the concentrations of the reactants and products remain unchanged over time. While it might seem that the reaction has stopped, both the forward and reverse reactions are still occurring, but at identical rates, resulting in no net change.

It is important to differentiate between chemical and physical changes when discussing equilibrium. In the context of gas laws, we often talk about physical changes (such as compression or expansion of a gas) that don’t involve chemical reactions. However, the concept of equilibrium is still relevant, as it's possible for physical processes to reach a state of balance too. For instance, when a gas is sealed in a container, it may eventually reach a point where its rate of expansion and compression balances out, even though no chemical reaction is taking place.

In the context of Boyle's Law problems, we must usually assume that the gas behaves ideally and that no chemical change affects the amount of gas present. If not, the pressure and volume changes could be influenced by reactions forming or consuming the gas, complicating the application of the law.
Aqueous Tension
Aqueous tension, often referred to as water vapor pressure, is the pressure exerted by water vapor present in a gas mixture or atmosphere. It becomes a significant factor to consider when collecting gases over water, as in the exercise provided. The presence of water vapor affects the total pressure exerted by the mixture of gases, because the water vapor itself takes up part of the pressure quota.

When solving gas law problems involving a gas collected over water, it's necessary to correct for aqueous tension to accurately determine the pressure of the desired gas. This is done by subtracting the aqueous tension from the total pressure measured, often given by a barometer. The resultant pressure is the actual pressure exerted by the gas in question. This correction is critical as it ensures we apply Boyle's Law accurately to just the gas of interest, without the confounding influence of water vapor.

Students' common mistake is overlooking this aspect, which leads to inaccurate calculations. Therefore, remembering to correct for aqueous tension not only improves understanding of gas behavior in wet conditions but also sharpens precision in problem-solving.

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Most popular questions from this chapter

The number of collisions of \(\mathrm{Ar}\) atoms with the walls of container per unit time (a) increases when the temperature increases. (b) remains the same when \(\mathrm{CO}_{2}\) is added to the container at constant temperature. (c) increases when \(\mathrm{CO}_{2}\) is added to the container at constant temperature. (d) decreases, when the average kinetic energy per molecule is decreased.

At which of the following four conditions, the density of an ideal gas will be maximum? (a) \(273 \mathrm{~K}\) and \(1 \mathrm{~atm}\) (b) \(273 \mathrm{~K}\) and \(2 \mathrm{~atm}\) (c) \(546 \mathrm{~K}\) and 1 atm (d) \(546 \mathrm{~K}\) and \(2 \mathrm{~atm}\)

Gases do not liquefy above the critical temperature because above critical temperature (a) the gases become ideal. (b) the intermolecular attraction vanishes. (c) the kinetic energy of molecules become so large that the attractive forces become unable to hold the molecules together. (d) the repulsive forces dominates in the molecules.

The volumes of the two vessels are in the ratio of \(2: 1\). One contains nitrogen and the other oxygen at \(800 \mathrm{~mm}\) and \(680 \mathrm{~mm}\) pressure, respectively. Determine the resulting pressure when they are connected together. (a) \(760 \mathrm{~mm}\) (b) \(670 \mathrm{~mm}\) (c) \(1140 \mathrm{~mm}\) (d) \(1480 \mathrm{~mm}\)

How many times the average speed of the molecules in a gas becomes when the temperature is raised from \(27^{\circ} \mathrm{C}\) to \(159^{\circ} \mathrm{C}\) ? (a) \(1.2\) (b) \(1.44\) (c) \(5.89\) (d) \(2.43\)

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