One mole of a monoatomic ideal gas undergoes a thermodynamic process such that`V^3/T^2`= constant. Then,

Options

A. work done by the gas is 200R when temperature of the gas is raised by 300 K
B. the specific heat of the gas in the process is `13/6R `
C. work done by the gas is 300R when temperature of the gas maised by 600 K
D. the specific heat of the gas in the process is `21/4` R

Answer

To solve the problem, we need to analyze the thermodynamic process described by the equation \( \frac{V^3}{T^2} = \text{constant} \). This indicates that the process is a polytropic process. We will derive the work done by the gas and the specific heat for this process step by step.

### Step-by-Step Solution:

1. **Identify the relationship**:
Given the equation \( \frac{V^3}{T^2} = C \), we can express \( T \) in terms of \( V \):
\[ T^2 = \frac{V^3}{C} \implies T = \sqrt{\frac{V^3}{C}} = \frac{V^{3/2}}{\sqrt{C}} \]

2. **Using the Ideal Gas Law**:
The ideal gas law states that \( PV = nRT \). For one mole of gas (\( n = 1 \)), we have:
\[ PV = RT \]
Substituting for \( T \) from the previous step:
\[ PV = R \cdot \frac{V^{3/2}}{\sqrt{C}} \implies P = \frac{R \cdot V^{1/2}}{\sqrt{C}} \]

3. **Finding the Polytropic Index \( x \)**:
In a polytropic process, \( PV^x = \text{constant} \). We can express \( P \) in terms of \( V \):
\[ P \propto V^{-x} \implies \frac{R \cdot V^{1/2}}{\sqrt{C}} \propto V^{-x} \]
This implies:
\[ V^{1/2 + x} = \text{constant} \]
Therefore, we have:
\[ 1/2 + x = 0 \implies x = -\frac{1}{2} \]

4. **Calculating the Work Done**:
The work done by the gas in a polytropic process is given by:
\[ W = \frac{nR \Delta T}{1 - x} \]
Here, \( n = 1 \) (one mole), and substituting \( x = -\frac{1}{2} \):
\[ W = \frac{R \Delta T}{1 - (-\frac{1}{2})} = \frac{R \Delta T}{1 + \frac{1}{2}} = \frac{R \Delta T}{\frac{3}{2}} = \frac{2R \Delta T}{3} \]

5. **Assuming a Change in Temperature**:
Let's assume a change in temperature \( \Delta T = 300 \, K \):
\[ W = \frac{2R \cdot 300}{3} = 200R \]

6. **Calculating Specific Heat**:
The specific heat \( C \) for a polytropic process is given by:
\[ C = \frac{R}{\gamma - 1} + \frac{R}{1 - x} \]
For a monoatomic gas, \( \gamma = \frac{5}{3} \):
\[ C = \frac{R}{\frac{5}{3} - 1} + \frac{R}{1 - (-\frac{1}{2})} = \frac{R}{\frac{2}{3}} + \frac{R}{\frac{3}{2}} = \frac{3R}{2} + \frac{2R}{3} \]
Finding a common denominator (6):
\[ C = \frac{9R}{6} + \frac{4R}{6} = \frac{13R}{6} \]

### Final Results:
- The work done by the gas is \( W = 200R \).
- The specific heat of the gas during this process is \( C = \frac{13R}{6} \).