If `c_(s)` is the velocity of sound in air and c is rms velocity , then

Options

A. `c_(s)ltc`
B. `c_(s)=c`
C. `c_(s)=c((gamma)/(3))^(1//2)`
D. None of these

Explanation

To solve the problem, we need to establish the relationship between the velocity of sound in air (\(c_s\)) and the root mean square (RMS) velocity (\(c\)). Let's break it down step by step.

### Step 1: Define the velocity of sound in air
The velocity of sound in air is given by the formula:
\[ c_s = \sqrt{\frac{\gamma P}{\rho}} \]
where:
- \(c_s\) = velocity of sound in air
- \(\gamma\) = adiabatic index (ratio of specific heats)
- \(P\) = pressure
- \(\rho\) = density of the air

### Step 2: Define the root mean square velocity
The root mean square velocity is defined as:
\[ c = \sqrt{\frac{3RT}{M}} \]
where:
- \(c\) = RMS velocity
- \(R\) = universal gas constant
- \(T\) = absolute temperature
- \(M\) = molar mass of the gas

### Step 3: Relate pressure, density, and temperature
From the ideal gas law, we know that:
\[ PV = nRT \]
For one mole of gas, this can be simplified to:
\[ P = \frac{RT}{V} \]
Also, the density \(\rho\) can be expressed as:
\[ \rho = \frac{m}{V} = \frac{M}{V} \]
Thus, we can relate pressure and density:
\[ P = \rho \frac{RT}{M} \]

### Step 4: Substitute the expression for pressure into the RMS velocity formula
Now, substituting \(P\) into the RMS velocity formula, we get:
\[ c = \sqrt{\frac{3P}{\rho}} = \sqrt{\frac{3 \cdot \rho \frac{RT}{M}}{\rho}} = \sqrt{\frac{3RT}{M}} \]

### Step 5: Relate \(c_s\) and \(c\)
Now we can express \(c_s\) in terms of \(c\):
\[ c_s = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma \cdot \rho \frac{RT}{M}}{\rho}} = \sqrt{\frac{\gamma RT}{M}} \]
From the earlier step, we have:
\[ c = \sqrt{\frac{3RT}{M}} \]
Thus, we can write:
\[ c_s = c \sqrt{\frac{\gamma}{3}} \]

### Final Result
So the relationship between the velocity of sound in air and the RMS velocity is:
\[ c_s = c \sqrt{\frac{\gamma}{3}} \]

### Summary
The final relationship derived is:
\[ c_s = c \sqrt{\frac{\gamma}{3}} \]