A sound wave passing through air at `NTP` produces a pressure of `0.001 "dyne"/cm^2` during a compression. The corresponding change in temperature (given `gamma = 1.5` and assume gas to be ideal) is

Options

A. `8.97 xx 10^(-4) K`
B. `9.87 xx 10^(-6) K`
C. `8.97 xx 10^(-8) K`
D. `9.87 xx 10^(-4) K`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Adiabatic Process
In an adiabatic process for an ideal gas, the relationship between pressure (P), volume (V), and temperature (T) is given by:
\[ PV^{\gamma} = \text{constant} \]
where \(\gamma\) (gamma) is the adiabatic exponent.

### Step 2: Relate Pressure and Temperature
From the ideal gas law, we know:
\[ PV = nRT \]
We can express \(V\) in terms of \(P\) and \(T\):
\[ V = \frac{nRT}{P} \]
Substituting this into the adiabatic condition gives:
\[ P \left(\frac{nRT}{P}\right)^{\gamma} = \text{constant} \]
This simplifies to:
\[ P^{1 - \gamma} T^{\gamma} = \text{constant} \]

### Step 3: Differentiate the Relationship
Taking the natural logarithm of both sides:
\[ \ln(P^{1 - \gamma}) + \ln(T^{\gamma}) = \ln(\text{constant}) \]
Differentiating gives:
\[ (1 - \gamma) \frac{dP}{P} + \gamma \frac{dT}{T} = 0 \]

### Step 4: Solve for Change in Temperature
Rearranging the equation:
\[ \gamma \frac{dT}{T} = -(1 - \gamma) \frac{dP}{P} \]
Thus,
\[ \frac{dT}{T} = -\frac{(1 - \gamma)}{\gamma} \frac{dP}{P} \]
Now, substituting \(\gamma = 1.5\):
\[ \frac{dT}{T} = -\frac{(1 - 1.5)}{1.5} \frac{dP}{P} = \frac{0.5}{1.5} \frac{dP}{P} \]

### Step 5: Substitute the Given Values
Given:
- \(dP = 0.001 \, \text{dyne/cm}^2\)
- At NTP, pressure \(P = 76 \, \text{cm of mercury} = 76 \times 13.6 \, \text{g/cm}^3 \times 981 \, \text{cm/s}^2\)

Calculating the atmospheric pressure in dyne/cm²:
\[ P = 76 \times 13.6 \times 981 \approx 1.013 \times 10^6 \, \text{dyne/cm}^2 \]

Now substituting these values into the equation:
\[ dT = \frac{0.5}{1.5} \cdot \frac{0.001}{1.013 \times 10^6} \cdot T \]

### Step 6: Calculate the Change in Temperature
At NTP, the temperature \(T\) is approximately \(273 \, \text{K}\):
\[ dT = \frac{0.5}{1.5} \cdot \frac{0.001}{1.013 \times 10^6} \cdot 273 \]

Calculating \(dT\):
\[ dT \approx \frac{0.5 \cdot 0.001 \cdot 273}{1.5 \cdot 1.013 \times 10^6} \]
\[ dT \approx \frac{0.1365}{1.5195 \times 10^6} \]
\[ dT \approx 8.97 \times 10^{-8} \, \text{K} \]

### Final Answer
The corresponding change in temperature is approximately:
\[ dT \approx 8.97 \times 10^{-8} \, \text{K} \]

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