Tempareture and volume of one mole of an ideal momatomic gas in a process are related as `TV^(2/3)=K` ,where k is constant.The molar specific heat capacity for the process is

Options

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

Answer

To find the molar specific heat capacity for the process described by the equation \( TV^{2/3} = K \), we can follow these steps:

### Step 1: Understand the relationship given
The relationship \( TV^{2/3} = K \) indicates a specific process involving temperature \( T \) and volume \( V \) of an ideal monatomic gas. Here, \( K \) is a constant.

### Step 2: Express temperature in terms of volume
From the equation, we can express temperature \( T \) as:
\[ T = \frac{K}{V^{2/3}} \]

### Step 3: Use the ideal gas law
According to the ideal gas law, we know that:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)), this simplifies to:
\[ PV = RT \]
Substituting our expression for \( T \) from Step 2 into the ideal gas law gives:
\[ PV = R \left( \frac{K}{V^{2/3}} \right) \]
Rearranging this, we can express \( P \) in terms of \( V \):
\[ P = \frac{RK}{V^{5/3}} \]

### Step 4: Identify the polytropic process
The relationship \( PV^{n} = \text{constant} \) identifies a polytropic process. From our expression for \( P \), we can see that:
\[ P \propto V^{-5/3} \]
Thus, we have:
\[ n = \frac{5}{3} \]

### Step 5: Calculate the molar specific heat capacity
The formula for molar specific heat capacity \( C \) for a polytropic process is given by:
\[ C = C_v + \frac{R}{1 - n} \]
For a monatomic ideal gas, the molar specific heat at constant volume \( C_v \) is:
\[ C_v = \frac{3R}{2} \]
Substituting \( n = \frac{5}{3} \) into the specific heat capacity formula:
\[ C = \frac{3R}{2} + \frac{R}{1 - \frac{5}{3}} \]
Calculating the denominator:
\[ 1 - \frac{5}{3} = -\frac{2}{3} \]
Thus:
\[ C = \frac{3R}{2} + \frac{R}{-\frac{2}{3}} = \frac{3R}{2} - \frac{3R}{2} = 0 \]

### Final Answer
The molar specific heat capacity for the process is:
\[ C = 0 \]