The root mean square speed of an ideal gas is given by : `u_("rms") = sqrt((3RT)/M)` Thus we conclude that `u_("rms")` speed of the ideal gas molecules is proportional to square root of the temperature and inversely proportional to the square root of the molar mass. The translational kinetic energy per mole can also be given as `1/2Mu_(rms)^(2)` . The mean free path (`lambda`) is the average of distances travelled by molecules in between two successive collisions whereas collision frequency (C.F.) is expressed as number of collisions taking place in unit time. The two terms `lambda` and C.F. are related by : `C.F = (u_("rms")/lambda)`
Which of the following relation is correct for an ideal gas regarding its pressure (P) and translational kinetic energy per unit volume (E) ?

Options

A. `P = 2/3 E`
B. `P = 3/2 E`
C. `P = 1/2 E`
D. P = 2E

Answer

To solve the question regarding the relationship between pressure (P) and translational kinetic energy per unit volume (E) for an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy of an Ideal Gas**:
The translational kinetic energy (KE) of an ideal gas can be expressed as:
\[ KE = \frac{1}{2} M u_{\text{rms}}^2 \]
where \(M\) is the mass of the gas molecules and \(u_{\text{rms}}\) is the root mean square speed.

2. **Relating Kinetic Energy to Pressure**:
The ideal gas law can be expressed in terms of kinetic energy. The pressure (P) exerted by the gas can be related to the average kinetic energy of the gas molecules. The relationship can be derived from the equation:
\[ PV = \frac{1}{3} N m u_{\text{rms}}^2 \]
where \(N\) is the number of molecules and \(m\) is the mass of a single molecule.

3. **Expressing Kinetic Energy per Unit Volume**:
The translational kinetic energy per unit volume (E) can be defined as:
\[ E = \frac{KE}{V} = \frac{1}{2} N m u_{\text{rms}}^2 \cdot \frac{1}{V} \]
From the ideal gas equation, we know that \(PV = nRT\), and for one mole of gas, \(n = \frac{N}{N_A}\) (where \(N_A\) is Avogadro's number).

4. **Substituting and Rearranging**:
We can substitute \(N\) in terms of \(P\) and \(V\):
\[ P = \frac{1}{3} N m u_{\text{rms}}^2 \implies N m = \frac{3PV}{u_{\text{rms}}^2} \]
Now substituting \(N m\) back into the equation for E:
\[ E = \frac{1}{2} \cdot \frac{3PV}{u_{\text{rms}}^2} \cdot u_{\text{rms}}^2 \cdot \frac{1}{V} = \frac{3P}{2} \]

5. **Final Relationship**:
Rearranging gives us the relationship between pressure and translational kinetic energy per unit volume:
\[ P = \frac{2}{3} E \]

Thus, the correct relation for an ideal gas regarding its pressure (P) and translational kinetic energy per unit volume (E) is:
\[ P = \frac{2}{3} E \]