The speed of sound in fresh water at `20^@C` is 1482 m/s. At what temperature is the speed of sound in helium gas the same as that of fresh water at `20^@` C? Helium is considered a monatomic ideal gas (`lamda` = 1.67 and atomic mass = 4.003 u).

Options

A. 313 K
B. 442 K
C. 633 K
D. 377 K

Answer

To solve the problem, we need to find the temperature at which the speed of sound in helium gas is equal to the speed of sound in fresh water at 20°C (which is 1482 m/s). We will use the formula for the speed of sound in an ideal gas:

\[ v = \sqrt{\frac{\gamma RT}{M}} \]

Where:
- \( v \) = speed of sound
- \( \gamma \) = adiabatic index (for helium, \( \gamma = 1.67 \))
- \( R \) = universal gas constant (\( R = 8.314 \, \text{J/(mol K)} \))
- \( T \) = absolute temperature in Kelvin
- \( M \) = molar mass of the gas in kg (for helium, \( M = 4.003 \, \text{g/mol} = 4.003 \times 10^{-3} \, \text{kg/mol} \))

### Step 1: Set up the equation
We know that the speed of sound in helium gas should equal the speed of sound in fresh water:

\[ 1482 = \sqrt{\frac{1.67 \times 8.314 \times T}{4.003 \times 10^{-3}}} \]

### Step 2: Square both sides
To eliminate the square root, we square both sides of the equation:

\[ 1482^2 = \frac{1.67 \times 8.314 \times T}{4.003 \times 10^{-3}} \]

### Step 3: Calculate \( 1482^2 \)
Calculating \( 1482^2 \):

\[ 1482^2 = 2197924 \]

### Step 4: Rearrange the equation
Now, rearranging the equation to solve for \( T \):

\[ T = \frac{1482^2 \times 4.003 \times 10^{-3}}{1.67 \times 8.314} \]

### Step 5: Substitute the values
Substituting the values into the equation:

\[ T = \frac{2197924 \times 4.003 \times 10^{-3}}{1.67 \times 8.314} \]

### Step 6: Calculate the numerator
Calculating the numerator:

\[ 2197924 \times 4.003 \times 10^{-3} = 8795.17 \]

### Step 7: Calculate the denominator
Calculating the denominator:

\[ 1.67 \times 8.314 = 13.88 \]

### Step 8: Final calculation for \( T \)
Now, substituting back into the equation for \( T \):

\[ T = \frac{8795.17}{13.88} \approx 633.0 \, \text{K} \]

### Conclusion
Thus, the temperature at which the speed of sound in helium gas is the same as that of fresh water at \( 20^\circ C \) is approximately \( 633 \, \text{K} \).

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