Mean kinetic energy of a perfect gas is proportional to

Options

A. `(1)/(T)`
B. `T^(@)`
C. `T^2`
D. T

Answer

To solve the question regarding the mean kinetic energy of a perfect gas and its proportionality, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Kinetic Energy in Gases**:
The kinetic energy (KE) of a perfect gas can be expressed in terms of the number of moles (n) and the temperature (T). The formula for the kinetic energy of n moles of a gas is given by:
\[ KE = \frac{3}{2} nRT \]
where \( R \) is the universal gas constant.

2. **Kinetic Energy for One Mole**:
To find the kinetic energy for one mole of gas, we can substitute \( n = 1 \) into the equation:
\[ KE = \frac{3}{2} RT \]

3. **Kinetic Energy Per Molecule**:
To find the kinetic energy per molecule, we need to divide the total kinetic energy of one mole by the number of molecules in one mole, which is Avogadro's number (\( N_A \)):
\[ KE_{\text{per molecule}} = \frac{KE}{N_A} = \frac{\frac{3}{2} RT}{N_A} \]

4. **Using the Boltzmann Constant**:
The term \( \frac{R}{N_A} \) is defined as the Boltzmann constant (\( k_B \)). Thus, we can rewrite the expression for kinetic energy per molecule as:
\[ KE_{\text{per molecule}} = \frac{3}{2} k_B T \]

5. **Conclusion on Proportionality**:
From the expression \( KE_{\text{per molecule}} = \frac{3}{2} k_B T \), we can see that the mean kinetic energy of a perfect gas is directly proportional to the absolute temperature \( T \). Therefore, we conclude that:
\[ KE \propto T \]

### Final Answer:
The mean kinetic energy of a perfect gas is proportional to the temperature of the gas.