In an adiabatic process, `R = (2)/(3) C_(v)`. The pressure of the gas will be proportional to:

Options

A. `T^(5//3)`
B. `T^(5//2)`
C. `T^(5//4)`
D. `T^(5//6)`

Explanation

To solve the problem step by step, we will analyze the relationship between pressure, temperature, and the specific heat capacities in an adiabatic process.

### Step 1: Understand the given relation
We start with the relation provided in the question:
\[ R = \frac{2}{3} C_v \]

### Step 2: Calculate \( C_v \)
From the given relation, we can express \( C_v \):
\[ C_v = \frac{3}{2} R \]

### Step 3: Use Mayer's relation to find \( C_p \)
Mayer's relation states:
\[ C_p - C_v = R \]
Substituting \( C_v \) into this equation gives:
\[ C_p - \frac{3}{2} R = R \]
Thus,
\[ C_p = R + \frac{3}{2} R = \frac{5}{2} R \]

### Step 4: Calculate the ratio \( \gamma \)
The ratio \( \gamma \) (gamma) is defined as:
\[ \gamma = \frac{C_p}{C_v} \]
Substituting the values we found:
\[ \gamma = \frac{\frac{5}{2} R}{\frac{3}{2} R} = \frac{5}{3} \]

### Step 5: Use the adiabatic relation
In an adiabatic process, the relationship between pressure \( P \), volume \( V \), and temperature \( T \) can be expressed as:
\[ T^\gamma P^{1 - \gamma} = \text{constant} \]
Rearranging this gives:
\[ P \propto T^{\frac{\gamma}{\gamma - 1}} \]

### Step 6: Substitute \( \gamma \) into the equation
Substituting \( \gamma = \frac{5}{3} \):
\[ P \propto T^{\frac{\frac{5}{3}}{\frac{5}{3} - 1}} \]
Calculating \( \gamma - 1 \):
\[ \frac{5}{3} - 1 = \frac{5}{3} - \frac{3}{3} = \frac{2}{3} \]
Thus,
\[ P \propto T^{\frac{\frac{5}{3}}{\frac{2}{3}}} = T^{\frac{5}{2}} \]

### Conclusion
Therefore, the pressure \( P \) of the gas is proportional to:
\[ P \propto T^{\frac{5}{2}} \]

### Final Answer
The pressure of the gas will be proportional to \( T^{\frac{5}{2}} \).

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