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At a given temperature internal energy o...

At a given temperature internal energy of a monoatomic, diatomic and non - linear gas is `(U_(0)` each. Find their translational and rotiational kinetic energies separately.

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To solve the problem, we need to find the translational and rotational kinetic energies for monoatomic, diatomic, and non-linear (polyatomic) gases given that their total internal energy is \( U_0 \). ### Step-by-Step Solution: 1. **Understanding Internal Energy**: The total internal energy \( U \) of a gas is related to its kinetic energy. For an ideal gas, the internal energy is the sum of translational and rotational kinetic energies. 2. **Monoatomic Gas**: - A monoatomic gas has only translational degrees of freedom. - The degrees of freedom for a monoatomic gas is 3 (all translational). - Therefore, the rotational kinetic energy is 0. - The translational kinetic energy can be calculated as: \[ K_{trans} = U_0 \] - Thus, for monoatomic gas: - Translational Kinetic Energy: \( K_{trans} = U_0 \) - Rotational Kinetic Energy: \( K_{rot} = 0 \) 3. **Diatomic Gas**: - A diatomic gas has 5 degrees of freedom: 3 translational and 2 rotational. - The translational kinetic energy is given by: \[ K_{trans} = \frac{3}{5} U_0 \] - The rotational kinetic energy is given by: \[ K_{rot} = \frac{2}{5} U_0 \] - Thus, for diatomic gas: - Translational Kinetic Energy: \( K_{trans} = \frac{3}{5} U_0 \) - Rotational Kinetic Energy: \( K_{rot} = \frac{2}{5} U_0 \) 4. **Non-linear (Polyatomic) Gas**: - A non-linear gas has 6 degrees of freedom: 3 translational and 3 rotational. - The translational kinetic energy is given by: \[ K_{trans} = \frac{3}{6} U_0 = \frac{1}{2} U_0 \] - The rotational kinetic energy is also: \[ K_{rot} = \frac{3}{6} U_0 = \frac{1}{2} U_0 \] - Thus, for non-linear gas: - Translational Kinetic Energy: \( K_{trans} = \frac{1}{2} U_0 \) - Rotational Kinetic Energy: \( K_{rot} = \frac{1}{2} U_0 \) ### Summary of Results: - **Monoatomic Gas**: - Translational Kinetic Energy: \( K_{trans} = U_0 \) - Rotational Kinetic Energy: \( K_{rot} = 0 \) - **Diatomic Gas**: - Translational Kinetic Energy: \( K_{trans} = \frac{3}{5} U_0 \) - Rotational Kinetic Energy: \( K_{rot} = \frac{2}{5} U_0 \) - **Non-linear Gas**: - Translational Kinetic Energy: \( K_{trans} = \frac{1}{2} U_0 \) - Rotational Kinetic Energy: \( K_{rot} = \frac{1}{2} U_0 \)

To solve the problem, we need to find the translational and rotational kinetic energies for monoatomic, diatomic, and non-linear (polyatomic) gases given that their total internal energy is \( U_0 \). ### Step-by-Step Solution: 1. **Understanding Internal Energy**: The total internal energy \( U \) of a gas is related to its kinetic energy. For an ideal gas, the internal energy is the sum of translational and rotational kinetic energies. 2. **Monoatomic Gas**: ...
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