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A flask with a volume of \(1.50 \mathrm{~L}\), provided with a stopcock, contains ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) at \(300 \mathrm{~K}\) and atmospheric pressure \(\left(1.013 \times 10^{5} \mathrm{~Pa}\right) .\) The molar mass of ethane is \(30.1 \mathrm{~g} / \mathrm{mol} .\) The system is warmed to a temperature of \(550 \mathrm{~K},\) with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

Short Answer

Expert verified
a) The final pressure of the ethane is calculated in step 3. b) The grams of ethane remaining in the flask are calculated in step 4.

Step by step solution

01

Calculate Initial Moles of Ethane

First, the initial amount of ethane before heating needs to be determined. This can be done by using the Ideal Gas Law equation, which is \[PV = nRT\], where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. Given \(P = 1.013 \times 10^{5} Pa\), \(V = 1.50 L\), \(R = 8.314 J/(mol \cdot K)\) (we use this value of \(R\) to match the units of pressure and volume given in the problem, and \(T = 300 K\), solve for \(n\).
02

Keep Track of Moles After Warming

The temperature changes to \(550 K\) with the stopcock open, implying that no change in the amount of gas since the ethane can escape to the atmosphere. This maintains the condition \(n = n_{initial}\).
03

Calculate Final Pressure of Ethane

The final pressure of ethane after cooling can be found by substituting the values of \(n\), \(R\), \(T\) (back to \(300 K\)), and \(V\) back into the ideal gas law equation and then solving for \(P\).
04

Calculate Mass of Ethane in Flask

To find the mass of ethane remaining in the flask, use the molar mass of ethane. Convert the number of moles calculated in step 1 (\(n\)) to mass by using the conversion factor \(1 mole = 30.1 g\). Use the formula \[Mass = n \times molar mass \]

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Key Concepts

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

Gas Constant
The gas constant, often symbolized as \( R \), is a crucial element in understanding the behavior of gases in different conditions. It appears in the Ideal Gas Law equation \( PV = nRT \), connecting pressure \( P \), volume \( V \), number of moles \( n \), and temperature in Kelvin \( T \). A key to using \( R \) correctly is to match its units with the other variables in the equation. For instance, in the given exercise, we use \( R = 8.314 J/(mol \cdot K) \) because the pressure is given in Pascals (Pa) and volume in liters (L).

Understanding the gas constant is essential for solving problems related to gas pressure, volume, and temperature, as it remains consistent for all ideal gases. This means that regardless of the type of the gas or the specific conditions it is under, the gas constant's value does not change, making it a powerful tool in carrying out calculations in chemistry and physics.

For those who aim to grasp the details of gas behavior, remembering the value of \( R \) and its proper application can be the difference between correct and incorrect solutions. It is notably used in scenarios that require a deeper comprehension of how a gas reacts when subjected to changes in conditions, such as temperature or volume variations.
Molar Mass
The molar mass is another fundamental concept deeply intertwined with stoichiometry and chemical calculations. It is defined as the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). In the context of the provided exercise, the molar mass of ethane (\( C_{2}H_{6} \)) is given as \( 30.1 g/mol \). This information is crucial in converting from moles to grams, which allows us to determine the actual mass of a gas present in a container.

Knowing the molar mass of a compound, like ethane in our example, helps us to not only link moles to grams but also to anticipate the mass of a gas that would occupy a standard volume at standard temperature and pressure. This is often a step in calculating the outcome of chemical reactions or the composition of gaseous mixtures. It also plays a significant role when comparing the behavior of different gases under the same conditions, as molar masses vary between different compounds.

Understanding molar mass becomes even more critical when dealing with complex problems where the gas's identity needs to be accounted for, and not just its behavior as an ideal gas.
Pressure-Temperature Relationship
The pressure-temperature relationship is part of the Ideal Gas Law and involves understanding how changes in temperature can affect the pressure of a gas when the volume is held constant. According to Gay-Lussac's Law, when a gas is heated or cooled at constant volume, the pressure will increase or decrease in direct proportion to the temperature. This means that if we increase the temperature, the pressure goes up, and if we decrease the temperature, the pressure goes down, assuming the amount of gas remains the same.

In the exercise, the flask is initially at a temperature of \( 300 K \), which increases to \( 550 K \). This change allows us to predict the behavior of the gas's pressure within the flask. When subsequently cooled back to its original temperature, we can deduce that the pressure will return to a value proportionate to the original temperature, provided the moles of gas remain steady.

This relationship illustrates why, under the assumption of a fixed volume, temperature serves as a fundamental variable in the determination of pressure. Without understanding this relationship, predicting and calculating the effects of temperature changes on gas pressure would be considerably more challenging.
Moles to Grams Conversion
Converting moles to grams is a crucial step in chemistry that allows us to move from the abstract world of moles, which relate to the number of particles, to the tangible world of mass that we can physically measure. To perform this conversion, we simply multiply the number of moles by the substance's molar mass. In mathematical terms, \( Mass = n \times molar mass \), where \( n \) is the number of moles.

In our exercise, after calculating the initial moles of ethane gas using the Ideal Gas Law, we can then use the molar mass of ethane to determine how many grams are present in the flask. Given that the molar mass of ethane is \( 30.1 g/mol \), the conversion is straightforward. This allows us to understand not just the volume and pressure that the gas occupies and exerts, but also the mass of the gas involved in the exercise. For many practical applications in both lab and industry settings, knowing the mass of a chemical reagent is critical for precise measurements and successful outcomes.

By mastering the skill of moles to grams conversion, students gain the ability to quantify the amount of substances in a chemical reaction or process, which forms a cornerstone of chemical quantitative analysis.

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Most popular questions from this chapter

A person at rest inhales \(0.50 \mathrm{~L}\) of air with each breath at a pressure of 1.00 atm and a temperature of \(20.0^{\circ} \mathrm{C}\). The inhaled air is \(21.0 \%\) oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of \(2000 \mathrm{~m}\) but the temperature is still \(20.0^{\circ} \mathrm{C}\). Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report "shortness of breath" at high elevations.

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains \(499 \mathrm{~cm}^{3}\) of air at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{~Pa}\right)\) and a temperature of \(27.0^{\circ} \mathrm{C}\). At the end of the stroke, the air has been compressed to a volume of \(46.2 \mathrm{~cm}^{3}\) and the gauge pressure has increased to \(2.72 \times 10^{6} \mathrm{~Pa}\). Compute the final temperature.

A pneumatic lift consists of a vertical cylinder with a radius of \(10.0 \mathrm{~cm} .\) A movable piston slides within the cylinder at its upper end and supports a platform on which loads are placed. An intake valve allows compressed air from a tank to enter the cylinder, and an exhaust valve allows air to be removed from the cylinder. In either case the rate of air transfer is sufficiently low that the temperature in the cylinder remains constant. When neither valve is activated, the cylinder is airtight. The piston and platform together have a mass of \(50.0 \mathrm{~kg},\) the temperature is \(20.0^{\circ} \mathrm{C},\) and the pressure outside the cylinder is 1.00 atm. (a) There is \(1.00 \mathrm{~mol}\) of air in the cylinder and no load on the platform. What is the height \(h\) between the bottom of the piston and the bottom of the cylinder? (b) A \(200 \mathrm{~kg}\) load is placed on the platform. By what distance does the platform drop? (c) The intake valve is activated and compressed air enters the cylinder so that the platform moves back to its original height. How many moles of air were introduced? (d) How many more moles of air should be introduced so that the platform rises \(2.00 \mathrm{~m}\) above its original height? (e) At what rate should this air be introduced, in \(\mathrm{mmol} / \mathrm{s}\), so that the platform rises at a rate of \(10.0 \mathrm{~cm} / \mathrm{s} ?\)

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