In a thermodynamic process two moles of a monatomic ideal gas obeys `PV^(–2)=constant`. If temperature of the gas increases from 300 K to 400 K, then find work done by the gas (Where R = universal gas constant)

Options

A. 200 R/3
B. `-200 R`
C. `-100R`
D. `-400R`

Answer

To solve the problem, we need to find the work done by the gas during a thermodynamic process where the gas follows the equation \( PV^{-2} = \text{constant} \). This indicates a polytropic process.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Number of moles, \( n = 2 \) moles.
- Initial temperature, \( T_1 = 300 \, K \).
- Final temperature, \( T_2 = 400 \, K \).
- Change in temperature, \( \Delta T = T_2 - T_1 = 400 \, K - 300 \, K = 100 \, K \).
- The polytropic index \( x = -2 \).

2. **Use the formula for work done in a polytropic process:**
The work done \( W \) in a polytropic process is given by:
\[ W = \frac{nR \Delta T}{1 - x} \]
where \( R \) is the universal gas constant.

3. **Substitute the values into the formula:**
\[ W = \frac{2R \cdot 100}{1 - (-2)} \]
Simplifying the denominator:
\[ 1 - (-2) = 1 + 2 = 3 \]
Thus, the equation becomes:
\[ W = \frac{200R}{3} \]

4. **Final result:**
The work done by the gas during the process is:
\[ W = \frac{200R}{3} \]

### Conclusion:
The work done by the gas is \( \frac{200R}{3} \).