In a process `V prop T^(2)`, temperature of 2 moles of a gas is increased by 200K. Work done by the gas in this process will be

Options

A. 600 R
B. 800 R
C. 1000 R
D. 1200 R

Answer

To solve the problem of finding the work done by the gas in the process where volume \( V \) is proportional to \( T^2 \), we can follow these steps:

### Step 1: Understand the relationship between volume and temperature
Given that \( V \propto T^2 \), we can express this as:
\[ V = kT^2 \]
where \( k \) is a constant.

### Step 2: Use the ideal gas law
From the ideal gas law, we know:
\[ PV = nRT \]
Substituting \( V \) from the previous step, we get:
\[ P(kT^2) = nRT \]
This simplifies to:
\[ P = \frac{nR}{k} \cdot \frac{1}{T} \]

### Step 3: Identify the polytropic process
In this case, since \( V \propto T^2 \), we can identify that this is a polytropic process with \( \gamma = \frac{1}{2} \).

### Step 4: Use the work done formula for a polytropic process
The work done \( W \) in a polytropic process is given by:
\[ W = \frac{nR \Delta T}{1 - \gamma} \]
where \( n \) is the number of moles, \( R \) is the gas constant, \( \Delta T \) is the change in temperature, and \( \gamma \) is the polytropic index.

### Step 5: Substitute the values
From the problem, we have:
- \( n = 2 \) moles
- \( \Delta T = 200 \, K \)
- \( \gamma = \frac{1}{2} \)

Substituting these values into the work done formula:
\[ W = \frac{2R \cdot 200}{1 - \frac{1}{2}} \]
\[ W = \frac{400R}{0.5} \]
\[ W = 800R \]

### Step 6: Conclusion
Thus, the work done by the gas in this process is:
\[ W = 800R \]