The kinetic energy of translation per unit volume of the molecules of hydrogen gas at N.T.P. is (Atmospheric pressure = `10%5N//m^2` )

Options

A. `1.5xx10^5J`
B. `1.5xx10^6J`
C. `1.25xx10^5J`
D. `1.25xx10^6J`

Answer

To find the kinetic energy of translation per unit volume of the molecules of hydrogen gas at N.T.P (Normal Temperature and Pressure), we can use the following steps:

### Step-by-Step Solution:

1. **Understand the relationship between kinetic energy and pressure**:
The kinetic energy (K) of N moles of gas can be expressed as:
\[ K = \frac{N F R T}{2} \]
where:
- \( N \) = number of moles
- \( F \) = degrees of freedom (for translation, \( F = 3 \))
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Relate the number of moles to pressure and volume**:
From the ideal gas law, we know:
\[ PV = NRT \]
This allows us to express \( NRT \) in terms of pressure (P) and volume (V):
\[ NRT = PV \]

3. **Substituting into the kinetic energy formula**:
We can substitute \( NRT \) in the kinetic energy formula:
\[ K = \frac{3PV}{2} \]

4. **Finding kinetic energy per unit volume**:
To find the kinetic energy per unit volume (\( K/V \)), we divide both sides by \( V \):
\[ \frac{K}{V} = \frac{3P}{2} \]

5. **Substituting the values**:
At N.T.P, the pressure \( P \) is given as \( 10^5 \, \text{N/m}^2 \):
\[ \frac{K}{V} = \frac{3 \times 10^5}{2} \]

6. **Calculating the final value**:
\[ \frac{K}{V} = \frac{3 \times 10^5}{2} = 1.5 \times 10^5 \, \text{J/m}^3 \]

### Final Answer:
The kinetic energy of translation per unit volume of the molecules of hydrogen gas at N.T.P is:
\[ 1.5 \times 10^5 \, \text{J/m}^3 \]