An ideal gas `(gamma = 1.5)` undergoes a thermodynamic process in which the temperature and pressure of the gas are related as `T^(-1)P^(2)`= constant. The molar heat capacity of the gas during the process is

Options

A. `(3)/(2)R`
B. `(4)/(3)R`
C. `(5)/(2)R`
D. `3R`

Answer

To find the molar heat capacity of the ideal gas undergoing the specified thermodynamic process, we can follow these steps:

### Step 1: Understand the relationship between temperature and pressure
We are given that \( T^{-1} P^2 = \text{constant} \). Let's denote this constant as \( k \). This can be rearranged to express pressure in terms of temperature:
\[ P^2 = k T \]
From this, we can express pressure \( P \) as:
\[ P = \sqrt{k T} \]

### Step 2: Use the ideal gas law
The ideal gas law states:
\[ PV = nRT \]
Substituting \( P \) from our previous expression into the ideal gas law gives:
\[ \sqrt{k T} V = nRT \]

### Step 3: Rearranging the equation
Rearranging the equation to isolate \( P \) gives us:
\[ P = \frac{nRT}{V} \]
Now, since we have \( P^2 \) in terms of \( T \), we can relate \( P \) and \( T \) through the volume \( V \).

### Step 4: Identify the type of process
From the relation \( P^2/T = \text{constant} \), we can identify that this is a polytropic process where:
\[ PV^x = \text{constant} \]
Comparing with the standard form, we find:
\[ x = -1 \]

### Step 5: Calculate the molar heat capacity
The formula for the molar heat capacity \( C \) during a polytropic process is given by:
\[ C = R \left( \frac{1}{\gamma - 1} + \frac{1}{1 - x} \right) \]
Substituting \( \gamma = 1.5 \) and \( x = -1 \):
\[ C = R \left( \frac{1}{1.5 - 1} + \frac{1}{1 - (-1)} \right) \]
Calculating each term:
\[ C = R \left( \frac{1}{0.5} + \frac{1}{2} \right) = R \left( 2 + 0.5 \right) = R \left( \frac{5}{2} \right) \]

### Step 6: Final result
Thus, the molar heat capacity of the gas during the process is:
\[ C = \frac{5R}{2} \]