\(f=\frac {2}{γ-1}\)
\(f=\frac {γ+1}{2}\)
\(f=\frac {2}{γ+1}\)
\(f=\frac {1}{γ+1}\)
To solve the given question, let's start by understanding the concept of degrees of freedom in the context of thermodynamics and kinetic theory of gases.
In the kinetic theory of gases, the degree of freedom refers to the number of independent ways a system can move without violating any constraints. For a monoatomic ideal gas, each atom can move in three perpendicular directions, so it has three translational degrees of freedom. Diatomic and polyatomic molecules have additional rotational and vibrational degrees of freedom.
The specific heat ratio (gamma, \( \gamma \)) is the ratio of the specific heat at constant pressure (\( C_p \)) to the specific heat at constant volume (\( C_v \)). It is given by:
\( \gamma = \frac{C_p}{C_v} \)
For an ideal gas, the degrees of freedom (\( f \)) is related to \( \gamma \) as follows:
\( f = \frac{2}{\gamma - 1} \)
This formula can be derived from the relation of the internal energy of a gas and its degrees of freedom. The internal energy (\( U \)) of an ideal gas is given by:
\( U = \frac{f}{2} nRT \)
Here, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
Now, let's justify why the correct answer is:
\( f = \frac{2}{\gamma - 1} \)
By using \( \gamma = \frac{C_p}{C_v} \), we know that:
\( C_p - C_v = R \)
\( \gamma = \frac{C_v + R}{C_v} \)
\( \gamma C_v = C_v + R \)
\( C_v (\gamma - 1) = R \)
Substitute this in the expression for degrees of freedom:
\( f = 3 - \frac{2C_v}{R} = 3 - \frac{2C_v}{C_v (\gamma - 1)} = 3 - \frac{2}{\gamma - 1} = \frac{2}{\gamma - 1} \)
Hence, the correct expression for the degrees of freedom is clearly \( f=\frac {2}{γ-1}\), which is option 1.
By eliminating the incorrect options: