The speed of sound through oxygen gas at T K is `v m s^(-1)`.As the temperature becomes 2T and oxygen gas dissociated into atomic oxygen, the speed of sound

Options

A. 1. remains the same
B. 2. become 2v
C. 3. become `sqrt2v`
D. 4. none of these

Answer

To solve the problem regarding the speed of sound through oxygen gas at different temperatures and states, we will follow these steps:

### Step 1: Understand the formula for the speed of sound
The speed of sound \( v \) in a gas is given by the formula:
\[ v = \sqrt{\frac{\gamma R T}{M}} \]
where:
- \( \gamma \) is the adiabatic index (ratio of specific heats),
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 2: Identify the initial conditions
At temperature \( T \):
- The speed of sound in oxygen gas is given as \( v \).

### Step 3: Analyze the change in conditions
When the temperature is increased to \( 2T \):
- The gas dissociates into atomic oxygen. This means we need to consider the change in the molar mass \( M \) and possibly the adiabatic index \( \gamma \).

### Step 4: Determine the new molar mass and adiabatic index
1. The molar mass of molecular oxygen \( O_2 \) is approximately \( 32 \, \text{g/mol} \).
2. When dissociated into atomic oxygen \( O \), the molar mass becomes \( 16 \, \text{g/mol} \).
3. The adiabatic index \( \gamma \) for diatomic gases (like \( O_2 \)) is typically around \( 1.4 \), while for monatomic gases (like \( O \)), it is \( 1.67 \).

### Step 5: Calculate the new speed of sound
Using the new conditions:
- At temperature \( 2T \), the speed of sound \( v' \) becomes:
\[ v' = \sqrt{\frac{\gamma' R (2T)}{M'}} \]
where \( \gamma' \) is the new adiabatic index for atomic oxygen and \( M' \) is the new molar mass.

Substituting the values:
\[ v' = \sqrt{\frac{1.67 R (2T)}{16}} \]

### Step 6: Relate the new speed to the original speed
We can express the new speed in terms of the original speed:
\[ v' = \sqrt{\frac{1.67 R (2T)}{16}} = \sqrt{\frac{1.67}{16}} \cdot \sqrt{2} \cdot \sqrt{\frac{R T}{M}} \]
Noting that \( v = \sqrt{\frac{\gamma R T}{M}} \):
\[ v' = \sqrt{\frac{1.67}{16}} \cdot \sqrt{2} \cdot v \]

### Step 7: Simplify the expression
To find the numerical value of \( v' \) in terms of \( v \):
- Calculate \( \sqrt{\frac{1.67}{16}} \) and \( \sqrt{2} \):
\[ v' = \sqrt{\frac{1.67 \cdot 2}{16}} \cdot v \]

### Conclusion
Thus, the new speed of sound \( v' \) is related to the original speed \( v \) by the factor derived above.