One kg of a diatomic gas is at pressure of `8xx10^4N//m^2`. The density of the gas is `4kg//m^3`. What is the energy of the gas due to its thermal motion?

Options

A. `3 xx 10^(4) J`
B. `5 xx 10^(4) J`
C. `6 xx 10^(4) J`
D. `7 xx 10^(4) J`

Answer

To find the energy of the gas due to its thermal motion, we can use the formula for the kinetic energy of a diatomic gas. The formula is given by:

\[ E = \frac{5}{2} nRT \]

However, in this problem, we are not directly given the number of moles (n), the gas constant (R), or the temperature (T). Instead, we are provided with the pressure (P) and density (ρ) of the gas.

### Step 1: Calculate the Volume of the Gas

We know that the density (ρ) is defined as:

\[ \rho = \frac{m}{V} \]

Where:
- \( m \) is the mass of the gas (1 kg)
- \( V \) is the volume of the gas

Rearranging this formula to find the volume gives us:

\[ V = \frac{m}{\rho} \]

Substituting the values:

\[ V = \frac{1 \text{ kg}}{4 \text{ kg/m}^3} = \frac{1}{4} \text{ m}^3 \]

### Step 2: Use the Formula for Energy in Terms of Pressure and Volume

The energy of the gas can also be expressed in terms of pressure (P) and volume (V):

\[ E = \frac{5}{2} PV \]

### Step 3: Substitute the Values of Pressure and Volume

Now, we can substitute the given pressure and the calculated volume into the energy formula:

\[ E = \frac{5}{2} \times (8 \times 10^4 \text{ N/m}^2) \times \left(\frac{1}{4} \text{ m}^3\right) \]

### Step 4: Simplify the Expression

Calculating the expression step-by-step:

1. Calculate \( P \times V \):

\[ P \times V = 8 \times 10^4 \times \frac{1}{4} = 2 \times 10^4 \text{ N m}^2 \]

2. Now substitute this back into the energy formula:

\[ E = \frac{5}{2} \times 2 \times 10^4 \]

3. Simplifying further:

\[ E = 5 \times 10^4 \text{ Joules} \]

### Final Answer

Thus, the energy of the gas due to its thermal motion is:

\[ E = 5 \times 10^4 \text{ Joules} \]

---