Problem 39 - Arihant AIEEE Physics

Question

In a model of chlorine \(\left(\mathrm{Cl}_{2}\right)\), two \(\mathrm{Cl}\) atoms are rotated about their centre of mass as shown. Here the two 'Cl' atoms are \(2 \times 10^{-10} \mathrm{~m}\) apart and angular speed \(\omega=2 \times 10^{12} \mathrm{rad} / \mathrm{s}\). If the molar mass of chlorine is \(70 \mathrm{~g} / \mathrm{mol}\), then what is the rotational kineti \(_{2}\) energy of one \(\mathrm{Cl}_{2}\) molecule? (a) \(2.32 \times 10^{-20} \mathrm{~J}\) (b) \(2.32 \times 10^{-21} \mathrm{~J}\) (c) \(2.32 \times 10^{-i>}\) \\} (d) \(2.32 \times 10^{-22} \mathrm{~J}\)

Step-by-Step Solution

Answer

The rotational kinetic energy of one \(\mathrm{Cl}_2\) molecule is approximately \(2.32 \times 10^{-21} \, \text{J}\), so the correct answer is (b).

Step 1: Calculate the mass of one Cl atom

To find the mass of one chlorine atom, remember that 1 mole of chlorine atoms has a mass of 70 g. Since Avogadro's number of atoms, which is approximately \(6.022 \times 10^{23}\), make up a mole, we can find the mass of a single chlorine atom by dividing the molar mass by Avogadro's number:\[m_{Cl} = \frac{70 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}}\]First, convert 70 g to kg, as SI units use kg:\[m_{Cl} = \frac{0.070 \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} = 1.162 \times 10^{-25} \, \text{kg/atom}\]

Step 2: Determine the moment of inertia for Cl2

The moment of inertia \(I\) for two atoms rotating about the center of mass is given by:\[I = m_{Cl} imes r^2\]where \(r\) is the distance of each Cl atom from the center of mass, which is half of the distance between the two atoms:\[r = \frac{2 \times 10^{-10} \, \text{m}}{2} = 1 \times 10^{-10} \, \text{m}\]Thus, the moment of inertia is:\[I = 1.162 \times 10^{-25} \, \text{kg} \times (1 \times 10^{-10} \, \text{m})^2\]\[I = 1.162 \times 10^{-45} \, \text{kg} \cdot \text{m}^2\]

Step 3: Calculate the rotational kinetic energy

The rotational kinetic energy \(K\) is given by the formula:\[K = \frac{1}{2} I \omega^2\]Plug in the values for \(I\) and \(\omega\):\[K = \frac{1}{2} \times 1.162 \times 10^{-45} \, \text{kg} \cdot \text{m}^2 \times (2 \times 10^{12} \, \text{rad/s})^2\]\[K = \frac{1}{2} \times 1.162 \times 10^{-45} \times 4 \times 10^{24}\]\[K = 2.324 \times 10^{-21} \, \text{J}\]

Step 4: Final Step: Select the correct answer

After calculating the rotational kinetic energy, compare your result to the provided choices. The calculated rotational kinetic energy is \(2.32 \times 10^{-21} \, \text{J}\). Thus, the correct answer is (b) \(2.32 \times 10^{-21} \, \text{J}\).

Key Concepts

Moment of Inertia

The moment of inertia is a crucial concept when studying rotational motion, akin to mass in linear motion. It measures how much torque is needed for a desired angular acceleration and depends on how the mass is distributed around the axis of rotation. For two chlorine atoms in a molecule, imagine them rotating about their center of mass. The moment of inertia, symbolized by \( I \), can be calculated using the formula: \[ I = m \cdot r^2 \] where \( m \) is the mass of one chlorine atom, and \( r \) is the distance from the atom to the axis of rotation. In this case, since chlorine atoms in \( Cl_2 \) are symmetrical around their center of mass, \( r \) is half the distance between the atoms. This calculation helps us determine how the molecule's mass distribution affects its rotational dynamics.

Angular Speed

Angular speed refers to how quickly an object rotates or revolves around a central point, measured in radians per second (rad/s). For our chlorine molecule problem, the angular speed, \( \omega \), is given as \( 2 \times 10^{12} \text{ rad/s} \). This property of motion enables us to understand how fast the \( Cl_2 \) molecule spins around its center of mass. Angular speed is the rotational equivalent of linear speed and is crucial in calculating rotational kinetic energy using the formula: \[ K = \frac{1}{2} I \omega^2 \] By knowing both the moment of inertia and angular speed, we can determine how energy is distributed within a spinning molecule or object.

Chlorine Molecule

A chlorine molecule, or \( Cl_2 \), consists of two chlorine atoms bonded together. These atoms are closely bound, making it a diatomic molecule. It is important in various chemical reactions and commonly used in disinfecting agents. In our context, imagine the chlorine atoms spinning around their mutual center of mass, creating rotational movement. Due to its symmetrical nature, each chlorine atom contributes equally to the molecule's moment of inertia. This symmetry simplifies calculations of rotational properties, aiding in exercises that explore the molecule's kinetic and dynamic behavior.

Molar Mass

Molar mass is a chemical property that represents the mass of one mole of a given substance. For chlorine, it's calculated as the sum of its atomic masses, resulting in \( 70 \text{ g/mol} \). This value is essential for calculating the mass of individual chlorine atoms, especially when converting to appropriate units for physics problems like finding the moment of inertia. Using molar mass, combined with Avogadro's number, we determine how much each atom weighs: \[ m_{Cl} = \frac{70 \thinspace \text{g/mol}}{6.022 \times 10^{23} \thinspace \text{atoms/mol}} \approx 1.162 \times 10^{-25} \thinspace \text{kg/atom} \] This conversion provides the necessary information for further calculations related to molecular dynamics.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry and physics, defining the number of atoms or molecules in one mole of a substance. It is approximately \( 6.022 \times 10^{23} \). This concept helps us understand the scale of individual atomic or molecular masses from bulk measurements like molar mass. By dividing the molar mass of chlorine by Avogadro's number, we can find the mass of a single chlorine atom, crucial for computing the moment of inertia and other physical properties in kinetic energy problems. The use of Avogadro's number bridges macroscopic measurements with microscopic scales, thereby making theoretical calculations practically meaningful.