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Containers \(\mathrm{A}\) and \(\mathrm{B}\) have same gases. Pressure, volume and temperature of \(\mathrm{A}\) are all twice that \(\mathrm{B}\), then the ratio of number of molecules of \(\mathrm{A}\) and \(\mathrm{B}\) are: (a) \(1: 2\) (b) 2 (c) \(1: 4\) (d) 4

Short Answer

Expert verified
The ratio of the number of molecules of A to B is 2.

Step by step solution

01

Understanding the Problem

We need to find the ratio of the number of molecules in container A to that in container B. Containers A and B contain the same gas, but the pressure, volume, and temperature in container A are each twice those in B. Use the ideal gas law to set up the relation.
02

Applying the Ideal Gas Law

The ideal gas law states that \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature. Let's apply this to both containers.
03

Setting Up Expressions for Each Container

For container B, the ideal gas law is \( P_B V_B = n_B R T_B \). For container A, since \( P_A = 2P_B \), \( V_A = 2V_B \), and \( T_A = 2T_B \), the equation becomes \( 2P_B \cdot 2V_B = n_A R \cdot 2T_B \).
04

Simplifying the Equation for Container A

Replacing values in container A: \( 4P_B V_B = n_A R \cdot 2T_B \). Simplify to \( 2P_B V_B = n_A R T_B \).
05

Equating Both Expressions

Now, equate the expressions from containers A and B: \( 2P_B V_B = n_A R T_B \) and \( P_B V_B = n_B R T_B \). Since both sides are equal (except for \( n_A \) and \( n_B \)), equate the moles for container A as double those for container B: \( 2n_B = n_A \).
06

Solve the Ratio of Number of Molecules

Since \( n_A = 2n_B \), the ratio of number of molecules \( n_A : n_B \) is 2:1. Thus, the ratio \( \text{number of molecules in A} : \text{number of molecules in B} \) is 2.

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Key Concepts

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

Pressure
Pressure is one of the fundamental concepts in understanding gases. It refers to the force exerted by gas particles on the walls of their container. Often, pressure in a gas is measured in units like Pascals (Pa) or atmospheres (atm).
Pressure is crucial in the ideal gas law, represented by the letter \( P \). In our given exercise, pressure in container A is twice that of container B, meaning that the force exerted by gas particles in A is substantially greater.
Imagine gas particles as tiny billiard balls constantly colliding with the container’s walls. More collisions mean more pressure. Since container A's pressure is doubled, it's like having twice as many billiard balls hitting the walls, leading to a higher force exerted per unit area.
Understanding how pressure interacts with other variables such as volume and temperature can help predict how gas will behave in different scenarios.
Volume
Volume is the space that the gas occupies in a container. It is usually measured in liters or cubic meters. Volume, represented by \( V \), is one of the pegs in the ideal gas law equation.
In the problem at hand, the volume of the gas in container A is twice that of container B. This means container A has much more space for gas particles to move around in than the gas in container B.
A simple way to visualize this is to picture two different-sized balloons. One is twice the size of the other. When a balloon is twice the volume, it can hold more particles or allow them more space to move, even if all other conditions, like pressure, are constant.
  • If the container's volume increases while everything else stays the same, the particles have more room, which impacts pressure and temperature.
Grasping how volume affects gas behavior is crucial for understanding how pressure, temperature, and moles of gas work together.
Temperature
Temperature is a measure of how hot or cold something is, relating directly to the kinetic energy of particles. It's measured in scales such as Celsius, Fahrenheit, or Kelvin, which is commonly used in scientific contexts.
In our scenario, the temperature of container A is twice that of container B. Temperature in gases affects how fast the particles are moving. Higher temperatures mean particles move more vigorously.
If we imagine gas particles are like miniature, hyperactive children bouncing around in a playroom, a higher temperature would mean they're moving faster and bumping into each other and the walls more frequently.
Temperature in the ideal gas law is symbolized as \( T \). The movement and energy of particles at different temperatures help us understand why gases behave differently under different conditions.
  • Higher temperatures lead to increased pressure if volume remains the same, as faster particles collide more forcefully with the container walls.
Thermal expansion and contraction, based on temperature changes, are key concepts in understanding gas behavior.
Moles of Gas
Moles of gas (represented by \( n \)) denote the amount of substance in a given container. A mole is a measure that lets us talk about the number of particles, in this case, gas molecules.
Our exercise compares the moles of gas between two containers. Container A, with its doubled pressure, volume, and temperature, ends up holding twice the number of moles compared to container B.
This relationship is clarified by the ideal gas law: \( PV = nRT \). When pressure, volume, and temperature in container A are each doubled, the amount of gas, or moles, is effectively doubled as well when solving for \( n \).
To visualize, think of moles as the count of invisible marbles inside a jar. Doubling the other factors means you're increasing the content of the jar without spilling marbles. This is why calculating the number of moles is crucial in stoichiometry, chemistry, and physics.
  • The mole concept helps in comprehending the proportionality among various components of gas.
  • Understanding moles allows for an accurate depiction of reactions and mixtures in gas forms.
Mastering this concept aids in deeper insights into how substances interact at the molecular level.

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Most popular questions from this chapter

Equal masses of methane and oxygen are mixed in an empty container at \(25^{\circ} \mathrm{C}\). The fraction of the total pressure exerted by oxygen is: (a) \(1 / 3 \times 273 / 298\) (b) \(1 / 3\) (c) \(1 / 2\) (d) \(2 / 3\)

In a mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\), the \(\mathrm{P}_{\mathrm{CO}_{2}}\) is \(0.4 \mathrm{~atm}\) and \(\mathrm{P}_{\text {Total }}\) is 2 atm. The percentage composition of the mixture by volume can be given as: (a) \(\mathrm{CO}_{2}=20 \%, \mathrm{H}_{2}=80 \%\) (b) \(\mathrm{CO}_{2}=40 \%, \mathrm{H}_{2}=60 \%\) (c) \(\mathrm{CO}_{2}=80 \%, \mathrm{H}_{2}=20 \%\) (d) \(\mathrm{CO}_{2}=60 \%, \mathrm{H}_{2}=40 \%\)

At a certain temperature for which \(\mathrm{RT}=25 \mathrm{~L}\) atm. \(\mathrm{mol}^{-1}\), the density of a gas, in \(\mathrm{g} \mathrm{L}^{-1}\), is \(\mathrm{d}=2.00 \mathrm{P}+\) \(0.020 \mathrm{P}^{2}\), where \(\mathrm{P}\) is the pressure in atmosphere. The molecular weight of the gas in \(\mathrm{g}\) mol- 1 is: (a) 60 (b) 75 (c) 50 (d) 35

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