A diatomic ideal gas is expanded according to `PV^(3)` = constant, under very high temperature (Assume vibration mode active). Calculate the molar heat capacity of gas (in cal / mol K) in this process.

Answer

To calculate the molar heat capacity of a diatomic ideal gas that is expanded according to the relation \( PV^{3} = \text{constant} \), we can follow these steps:

### Step 1: Identify the parameters of the gas
Given that the gas is diatomic and at very high temperatures, we need to consider the degrees of freedom. For a diatomic gas, the degrees of freedom include translational, rotational, and vibrational modes.

At high temperatures, the vibrational modes become active, which means we need to account for them in our calculations.

### Step 2: Determine the molar heat capacity at constant volume (\( C_V \))
For a diatomic gas, the molar heat capacity at constant volume (\( C_V \)) can be calculated as follows:

\[ C_V = \frac{5}{2} R + R_{vib} \]

Where:
- \( R \) is the universal gas constant.
- \( R_{vib} \) is the contribution from the vibrational modes.

At high temperatures, each vibrational mode contributes \( R \) to the heat capacity. A diatomic molecule has one vibrational mode, so:

\[ R_{vib} = R \]

Thus, we have:

\[ C_V = \frac{5}{2} R + R = \frac{7}{2} R \]

### Step 3: Calculate the molar heat capacity in the process
The given process follows the relation \( PV^{3} = \text{constant} \). In this case, we can use the relation for molar heat capacity in a polytropic process:

\[ C = C_V + R \left(1 - \frac{1}{n}\right) \]

Where \( n \) is the polytropic index. From the relation \( PV^{3} = \text{constant} \), we can identify \( n = 3 \).

Substituting \( C_V \) and \( n \):

\[ C = \frac{7}{2} R + R \left(1 - \frac{1}{3}\right) \]

Calculating \( R \left(1 - \frac{1}{3}\right) \):

\[ R \left(1 - \frac{1}{3}\right) = R \left(\frac{2}{3}\right) = \frac{2}{3} R \]

Now substituting back into the equation for \( C \):

\[ C = \frac{7}{2} R + \frac{2}{3} R \]

### Step 4: Finding a common denominator and simplifying
To add these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6:

\[ C = \frac{21}{6} R + \frac{4}{6} R = \frac{25}{6} R \]

### Step 5: Convert to calories per mole per Kelvin
Using \( R = 2 \) cal/(mol K) (the value of the gas constant in calories):

\[ C = \frac{25}{6} \times 2 = \frac{50}{6} \text{ cal/(mol K)} \approx 8.33 \text{ cal/(mol K)} \]

### Final Answer
The molar heat capacity of the gas in this process is approximately \( 8.33 \text{ cal/(mol K)} \).

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