Kinetic energy and pressure of a gas of unit volume are related as:

Options

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

Explanation

To solve the problem of how kinetic energy and pressure of a gas of unit volume are related, we can follow these steps:

### Step 1: Understand the Kinetic Energy of a Gas
The kinetic energy (E) of an ideal gas can be expressed as:
\[ E = \frac{3}{2} nRT \]
where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = absolute temperature in Kelvin

### Step 2: Relate Pressure to Volume
For an ideal gas, the equation of state is given by:
\[ PV = nRT \]
Since we are considering a unit volume (V = 1), we can simplify this to:
\[ P = nRT \]

### Step 3: Substitute Pressure into the Kinetic Energy Equation
Now, we can substitute the expression for pressure (P) into the kinetic energy equation. From Step 2, we have:
\[ E = \frac{3}{2} nRT \]
Substituting \( nRT \) from the pressure equation:
\[ E = \frac{3}{2} P \]

### Step 4: Rearranging the Equation
From the equation \( E = \frac{3}{2} P \), we can rearrange it to express pressure in terms of kinetic energy:
\[ P = \frac{2}{3} E \]

### Conclusion
Thus, the relationship between the kinetic energy and pressure of a gas of unit volume is:
\[ P = \frac{2}{3} E \]

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