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Suppose you have two 1 -L flasks, one containing \(\mathrm{N}_{2}\) at \(\mathrm{STP}\), the other containing \(\mathrm{CH}_{4}\) at STP. How do these systems \(\mathrm{com}\) pare with respect to (a) number of molecules, (b) density, (c) average kinetic energy of the molecules, \((\mathbf{d})\) rate of effusion through a pinhole leak?

Short Answer

Expert verified
(a) The number of molecules in both flasks is the same, as they contain equal moles of gas at STP. (b) The density of N2 is higher than the density of CH4, as its molar mass is greater. (c) The average kinetic energy of the molecules is equal for both gases, as they are at the same temperature. (d) The rate of effusion of CH4 is greater than the rate of effusion of N2, as its molar mass is lower.

Step by step solution

01

a) Number of molecules

At STP, the pressure is 1 atm and the temperature is 273.15 K. To find the number of molecules, we will use the Ideal Gas Law Equation. PV = nRT Here, P = 1 atm, T = 273.15 K, R (gas constant) = 0.0821 atm L/mol K For N2: Molar Mass of N2 = 28 g/mol For CH4: Molar Mass of CH4 = 16 g/mol First, we need to find the moles of each gas in the flask, then multiply by Avogadro's number to find the number of molecules. n_N2 = (PV) / (RT) n_CH4 = (PV) / (RT) Based on the values of n_N2 and n_CH4, we will then find the number of molecules.
02

b) Density

Density of a gas is given by: Density = Mass / Volume Density_N2 = (Mass of N2) / (Volume of flask) Density_CH4 = (Mass of CH4) / (Volume of flask) We can find the mass of each gas using the moles and molar mass, then divide by volume (1 L) to obtain the densities of N2 and CH4.
03

c) Average kinetic energy of the molecules

The average kinetic energy of a gas depends on temperature alone and is given by: KE_avg = (3/2) * R * T Since both gases are at the same temperature (STP), their average kinetic energies are equal.
04

d) Rate of effusion through a pinhole leak

Rate of effusion is given by Graham's Law of Effusion: Rate1 / Rate2 = sqrt(MolarMass2 / MolarMass1) Here, Rate1 refers to the rate of effusion of N2, and Rate2 refers to the rate of effusion of CH4. Using the molar masses of N2 and CH4, we can calculate the rate of effusion through a pinhole leak for both gases and compare them.

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Key Concepts

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

molecular comparison
When comparing gases at Standard Temperature and Pressure (STP), the number of molecules in a container depends on the conditions defined by the Ideal Gas Law. The formula \( PV = nRT \) helps us determine the moles of gas present. Using this equation, we can find that at STP, 1 mole of any gas occupies 22.4 liters.Therefore, each 1-liter flask of \(N_2\) and \(CH_4\) contains the same number of moles. Since both gases are in equal volume and under identical conditions, they have an equal number of molecules. This happens because of Avogadro's principle, which states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules.Hence, despite \(N_2\) and \(CH_4\) having different molecular structures and molar masses, their amount in moles and, therefore, the number of molecules are equivalent in their respective flasks.
density of gases
Density is a measure of how much mass is contained in a given volume. The formula for the density of a gas is generally given by the equation:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]To determine the density of \(N_2\) and \(CH_4\) under the same conditions, we need to consider their molar masses. - For \(N_2\), the molar mass is 28 g/mol.- For \(CH_4\), the molar mass is 16 g/mol.Because both gases occupy a 1-liter volume at STP, we can use these molar masses to calculate their densities. Multiply the number of moles of each gas (at STP, 1 mole in 22.4 liters) by the respective molar mass, and then divide by the flask volume.Consequently, the density of \(CH_4\) is less than that of \(N_2\), due to its smaller molar mass. This explains why lighter gases have lower densities when compared to heavier gases under the same conditions.
average kinetic energy
The average kinetic energy of gas molecules is directly related to temperature, irrespective of the type of gas or its molar mass. It is given by the expression:\[ KE_{\text{avg}} = \frac{3}{2} k_B T \]where \(k_B\) is Boltzmann's constant and \(T\) is the temperature. Because both \(N_2\) and \(CH_4\) are at STP, they are at the same temperature, leading to identical average kinetic energies of their molecules.This concept emphasizes that temperature, not the type or mass of gas molecules, dictates the kinetic energy. Therefore, any gas at a given temperature (in an ideal setup) will have molecules with the same average kinetic energy. Thus, understand that while the masses of \(N_2\) and \(CH_4\) molecules differ, their movements at the same temperature are just as energetically active.
Graham's Law of Effusion
Graham's Law of Effusion describes how different gases diffuse or effuse through small openings, such as a pinhole. It is mathematically stated as:\[ \frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{Molar mass_{2}}{Molar mass_{1}}} \]Given the problem setup, for \(N_2\) and \(CH_4\), we will compare their rates of effusion. Since \(N_2\) has a greater molar mass (28 g/mol) compared to \(CH_4\) (16 g/mol), \(CH_4\) will effuse faster.Graham's Law highlights that lighter gases effuse more quickly than heavier ones due to their lower molar masses. This property is why gases like helium and hydrogen, being very light, diffuse through materials at a faster rate than heavier gases.

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Most popular questions from this chapter

Ammonia and hydrogen chloride react to form solid ammonium chloride: $$\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)$$ Two \(2.00-\mathrm{L}\) flasks at \(25^{\circ} \mathrm{C}\) are connected by a valve, as shown in the drawing. One flask contains \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}(g),\) and the other contains \(5.00 \mathrm{~g}\) of \(\mathrm{HCl}(g) .\) When the valve is opened, the gases react until one is completely consumed. (a) Which gas will remain in the system after the reaction is complete? (b) What will be the final pressure of the system after the reaction is complete? (Neglect the volume of the ammonium chloride formed.) (c) What mass of ammonium chloride will be formed?

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathbf{a})\) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at 99.7 \(\mathrm{kPa}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise 10.53 . The unknown vapor had a mass of \(2.55 \mathrm{~g} ;\) the volume of the bulb was \(500 \mathrm{~mL}\), pressure \(101.33 \mathrm{kPa}\), and temperature \(37^{\circ} \mathrm{C.Calculate}\) the molar mass of the unknown vapor.

(a) If the pressure exerted by ozone, \(\mathrm{O}_{3}\), in the stratosphere is \(304 \mathrm{~Pa}\) and the temperature is \(250 \mathrm{~K}\), how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately \(0.04 \%\) of Earth's atmosphere. If you collect a 2.0-L sample from the atmosphere at sea level \((101.33 \mathrm{kPa})\) on a warm day \(\left(27^{\circ} \mathrm{C}\right)\), how many \(\mathrm{CO}_{2}\) molecules are in your sample?

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{O}_{2}, \mathrm{Ar}, \mathrm{CO}, \mathrm{HCl}, \mathrm{CH}_{4} \cdot(\mathbf{b})\) Calculate the rms speed of CO molecules at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) Calculate the most probable speed of an argon atom in the stratosphere, where the temperature is \(0^{\circ} \mathrm{C}\).

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