A perfect gas undergoes a reversible adiabatic expansion from (300 K, 200 atm) to (90 K, 10 atm). Find the atomicity of gas.

Answer

To solve the problem of finding the atomicity of a perfect gas undergoing a reversible adiabatic expansion from (300 K, 200 atm) to (90 K, 10 atm), we will follow these steps:

### Step 1: Understand the Adiabatic Process
In a reversible adiabatic process, the relationship between pressure (P), volume (V), and temperature (T) can be described using the equation:
\[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \]
where \( \gamma \) (gamma) is the heat capacity ratio, defined as:
\[ \gamma = \frac{C_p}{C_v} \]

### Step 2: Use the Temperature-Pressure Relationship
For an adiabatic process, we can also use the relationship:
\[ \frac{P_1}{P_2} = \left( \frac{T_2}{T_1} \right)^{\frac{\gamma}{\gamma - 1}} \]

### Step 3: Substitute the Given Values
We have:
- Initial temperature, \( T_1 = 300 \, K \)
- Final temperature, \( T_2 = 90 \, K \)
- Initial pressure, \( P_1 = 200 \, atm \)
- Final pressure, \( P_2 = 10 \, atm \)

Substituting these values into the equation:
\[ \frac{200}{10} = \left( \frac{90}{300} \right)^{\frac{\gamma}{\gamma - 1}} \]

### Step 4: Simplify the Equation
This simplifies to:
\[ 20 = \left( \frac{3}{10} \right)^{\frac{\gamma}{\gamma - 1}} \]

### Step 5: Take Logarithm of Both Sides
Taking the logarithm of both sides gives us:
\[ \log(20) = \frac{\gamma}{\gamma - 1} \log\left(\frac{3}{10}\right) \]

### Step 6: Express Logarithms
Using properties of logarithms, we can express:
\[ \log(20) = \log(2) + \log(10) = \log(2) + 1 \]
And:
\[ \log\left(\frac{3}{10}\right) = \log(3) - \log(10) = \log(3) - 1 \]

### Step 7: Substitute Logarithmic Values
Substituting these into the equation gives:
\[ \log(2) + 1 = \frac{\gamma}{\gamma - 1} (\log(3) - 1) \]

### Step 8: Solve for Gamma
Rearranging and solving for \( \gamma \) involves some algebraic manipulation. We can express it as:
\[ \frac{\log(2) + 1}{\log(3) - 1} = \frac{\gamma}{\gamma - 1} \]

### Step 9: Calculate Values
Using approximate values:
- \( \log(2) \approx 0.301 \)
- \( \log(3) \approx 0.477 \)

Substituting these values gives:
\[ \frac{0.301 + 1}{0.477 - 1} = \frac{\gamma}{\gamma - 1} \]

### Step 10: Find Atomicity
After solving for \( \gamma \), we can determine the atomicity of the gas:
- For monoatomic gases, \( \gamma = \frac{5}{3} \)
- For diatomic gases, \( \gamma = \frac{7}{5} \)
- For polyatomic gases, \( \gamma < \frac{5}{3} \)

Given that \( \gamma \) calculated is approximately \( 1.66 \), which is close to \( \frac{5}{3} \), we conclude that the gas is monoatomic.

### Final Answer
The atomicity of the gas is **1** (monoatomic).

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