If the temperature is raised by 1 K from 300 K, the percentage change in the speed of sound in the gaseous mixture is `[R=8.314" J mol"^(-1)K^(-1)]`

Options

A. (a)`0.167%`
B. (b)`0.334%`
C. (c)`1%`
D. (d)`2%`

Answer

To solve the problem of finding the percentage change in the speed of sound in a gaseous mixture when the temperature is raised by 1 K from 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Speed of Sound:**
The speed of sound \( V_s \) in a gas can be expressed as:
\[ V_s = \sqrt{\frac{\gamma R T}{M}} \]
where:
- \( \gamma \) is the adiabatic constant (ratio of specific heats),
- \( R \) is the gas constant (8.314 J/mol·K),
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

2. **Calculate the Initial Speed of Sound:**
At \( T = 300 \, K \):
\[ V_{s1} = \sqrt{\frac{\gamma R \cdot 300}{M}} \]

3. **Calculate the Speed of Sound After Temperature Increase:**
When the temperature is raised by 1 K (to 301 K):
\[ V_{s2} = \sqrt{\frac{\gamma R \cdot 301}{M}} \]

4. **Find the Change in Speed of Sound:**
The change in speed of sound \( \Delta V_s \) can be approximated using the formula for small changes:
\[ \Delta V_s = V_{s2} - V_{s1} \]

5. **Use the Approximation for Small Changes:**
For small changes in temperature, we can use the derivative approximation:
\[ \frac{\Delta V_s}{V_s} \approx \frac{1}{2} \frac{\Delta T}{T} \]
Here, \( \Delta T = 1 \, K \) and \( T = 300 \, K \).

6. **Substitute Values:**
\[ \frac{\Delta V_s}{V_s} \approx \frac{1}{2} \cdot \frac{1}{300} \]

7. **Calculate the Percentage Change:**
To find the percentage change:
\[ \frac{\Delta V_s}{V_s} \cdot 100\% \approx \frac{1}{2} \cdot \frac{1}{300} \cdot 100\% \]
\[ = \frac{100}{600} = \frac{1}{6} \approx 0.1667\% \]

### Final Answer:
The percentage change in the speed of sound when the temperature is raised by 1 K from 300 K is approximately \( 0.1667\% \).