The temperature of n moles of an ideal gas is increased from T to 4T through a process for which pressure `P=aT^(-1)` where a is a constant .Then the work done by the gas is

Options

A. a. nRT
B. b. 4nRT
C. c. 2nRT
D. d. 6nRT

Explanation

To solve the problem, we need to calculate the work done by the gas when the temperature of n moles of an ideal gas is increased from T to 4T under the given pressure condition \( P = aT^{-1} \).

### Step-by-Step Solution:

1. **Identify the Work Done Formula**:
The work done \( dW \) by the gas during an infinitesimal change in volume is given by:
\[ dW = P \, dV \]

2. **Substitute for Pressure**:
We know from the problem that the pressure \( P \) is given by:
\[ P = aT^{-1} \]

3. **Use Ideal Gas Law**:
The ideal gas law states:
\[ PV = nRT \]
Rearranging this gives us the volume \( V \):
\[ V = \frac{nRT}{P} \]
Substituting for \( P \):
\[ V = \frac{nRT}{aT^{-1}} = \frac{nRT^2}{a} \]

4. **Differentiate Volume**:
To find \( dV \), we differentiate \( V \) with respect to \( T \):
\[ dV = \frac{d}{dT}\left(\frac{nRT^2}{a}\right) = \frac{2nRT}{a} \, dT \]

5. **Substitute \( dV \) into Work Done**:
Now substitute \( dV \) back into the work done formula:
\[ dW = P \, dV = aT^{-1} \cdot \frac{2nRT}{a} \, dT \]
The \( a \) cancels out:
\[ dW = 2nR \, dT \]

6. **Integrate to Find Total Work Done**:
Now we integrate \( dW \) from the initial temperature \( T \) to the final temperature \( 4T \):
\[ W = \int_{T}^{4T} 2nR \, dT = 2nR \int_{T}^{4T} dT \]
This evaluates to:
\[ W = 2nR \left[ T \right]_{T}^{4T} = 2nR (4T - T) = 2nR (3T) = 6nRT \]

### Final Answer:
Thus, the work done by the gas is:
\[ W = 6nRT \]