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Chapter 10: Problem 3

Consider the sample of gas depicted here. What would the drawing look like if the volume and temperature remained constant while you removed enough of the gas to decrease the pressure by a factor of 2\(?[\) Section 10.3\(]\) \begin{equation}\begin{array}{l}{\text { (a) It would contain the same number of molecules. }} \\ {\text { (b) It would contain half as many molecules. }} \\\ {\text { (c) It would contain twice as many molecules. }} \\ {\text { (d) There is insufficient data to say. }}\end{array}\end{equation}

Short Answer

Expert verified
(b) It would contain half as many molecules.

Step by step solution

01

Apply Boyle's Law

As the volume V and the temperature T remain constant while the pressure P changes, we can apply Boyle's Law. Boyle's Law states that the product of the pressure and volume (P 脳 V) remains constant for an ideal gas when held at a constant temperature. Mathematically, it can be represented as: \(P_1V_1 = P_2V_2\) where \(P_1\) and \(P_2\) are the initial and final pressures, and \(V_1\) and \(V_2\) are the initial and final volumes.
02

Use the given condition to find the relationship between the initial and final pressures

Given that the pressure decreases by a factor of 2, we can relate the initial and final pressure as: \(P_2 = \dfrac{P_1}{2}\) Since the volume remains constant, we can say that \(V_1 = V_2\).
03

Determine the relationship between the initial and final number of molecules

Using the Ideal Gas Law, we can find the relationship between the initial and final number of molecules of the gas. The Ideal Gas Law is expressed as: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. For the initial state: \(P_1V_1 = n_1RT_1\) For the final state: \(P_2V_2 = n_2RT_2\) As temperature and volume remain constant, we have: \(T_1 = T_2\) and \(V_1 = V_2\). Since \(P_1V_1 = P_2V_2\), we can write: \(n_1RT_1 = n_2RT_2\) Dividing both sides by RT: \(\dfrac{n_1T_1}{n_2T_2} = \dfrac{n_1}{n_2}\) As T is constant, we have: \(\dfrac{n_1}{n_2} = \dfrac{P_1V_1}{P_2V_2}\) Since the volume is constant, we can write the relation as: \(\dfrac{n_1}{n_2} = \dfrac{P_1}{P_2}\) Now substituting the value of \(P_2\) from step 2: \(\dfrac{n_1}{n_2} = \dfrac{P_1}{\frac{P_1}{2}}\) \(\dfrac{n_1}{n_2} = 2\) Here, \(n_2 = \dfrac{n_1}{2}\) which means there are now half as many molecules.
04

Choose the correct answer based on the analysis

Based on the relationship between the initial and final number of molecules, the correct answer is: (b) It would contain half as many molecules.

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Most popular questions from this chapter

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 \(0.846 \mathrm{~g}\); the volume of the bulb was \(354 \mathrm{~cm}^{3}\), pressure 752 torr, and temperature \(100^{\circ} \mathrm{C}\). Calculate the molar mass of the unknown vapor.

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? (a) Because Xe is a noble gas, there is less tendency for the Xe atoms to repel one another, so they pack more densely in the gaseous state. (b) Xe atoms have a higher mass than \(\mathrm{N}_{2}\) molecules. Because both gases at STP have the same number of molecules per unit volume, the Xe gas must be denser. (c) The Xe atoms are larger than \(\mathrm{N}_{2}\) molecules and thus take up a larger fraction of the space occupied by the gas. (d) Because the Xe atoms are much more massive than the \(\mathrm{N}_{2}\) molecules, they move more slowly and thus exert less upward force on the gas container and make the gas appear denser.

Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations. Consider a 1000 -megawatt coal-fired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, \(1.00 \mathrm{~atm}\), and \(27^{\circ} \mathrm{C}\), calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. (b) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and \(120 \mathrm{~atm}\) and a density of \(1.2 \mathrm{~g} / \mathrm{cm}^{3}\), what volume does it possess? (c) If it is stored underground as a gas at \(36^{\circ} \mathrm{C}\) and \(90 \mathrm{~atm}\), what volume does it occupy?

Which of the following statements best explains why a closed balloon filled with helium gas rises in air? (a) Helium is a monatomic gas, whereas nearly all the molecules that make up air, such as nitrogen and oxygen, are diatomic. (b) The average speed of helium atoms is greater than the average speed of air molecules, and the greater speed of collisions with the balloon walls propels the balloon upward. (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The mass of the balloon is thus less than the mass of the air displaced by its volume. (d) Because helium has a lower molar mass than the average air molecule, the helium atoms are in faster motion. This means that the temperature of the helium is greater than the air temperature. Hot gases tend to rise.

The accompanying drawing represents a mixture of three different gases. (a) Rank the three components in order of increasing partial pressure. (b) If the total pressure of the mixture is \(1.40\) atm, calculate the partial pressure of each gas. [Section 10.6]
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