n moles of an ideal gas undergo a process in which the temperature changes with volume as `T=kv^(2)`. The work done by the gas as the temperature changes from `7_(0)` to `47_(0)` is

Options

A. `3nRT_(0)`
B. `((5)/(2))nRT_(0)`
C. `((3)/(2))nRT_(0)`
D. Zero

Explanation

To solve the problem of finding the work done by an ideal gas as the temperature changes from \(7^\circ C\) to \(47^\circ C\) with the relationship \(T = kV^2\), we can follow these steps:

### Step 1: Identify the relationship between temperature and volume
Given the equation \(T = kV^2\), we can express the volume in terms of temperature:
\[ V = \sqrt{\frac{T}{k}} \]

### Step 2: Differentiate the temperature with respect to volume
To find the work done, we need to express the work done \(dW\) in terms of volume. The differential form of the temperature is:
\[ dT = 2kVdV \]
From this, we can express \(dV\) in terms of \(dT\):
\[ dV = \frac{dT}{2kV} \]

### Step 3: Use the ideal gas law
According to the ideal gas law, we have:
\[ PV = nRT \]
From the relationship \(T = kV^2\), we can substitute \(T\) into the ideal gas law:
\[ P = \frac{nR(kV^2)}{V} = nRkV \]

### Step 4: Express work done \(dW\)
The work done by the gas during an infinitesimal expansion is given by:
\[ dW = PdV = nRkVdV \]
Substituting \(dV\) from the previous step:
\[ dW = nRkV \left(\frac{dT}{2kV}\right) = \frac{nR}{2} dT \]

### Step 5: Integrate to find total work done
Now we need to integrate \(dW\) from the initial temperature \(T_1 = 7^\circ C\) to the final temperature \(T_2 = 47^\circ C\):
\[ W = \int_{T_1}^{T_2} \frac{nR}{2} dT = \frac{nR}{2} (T_2 - T_1) \]
Substituting the values:
\[ W = \frac{nR}{2} (47 - 7) = \frac{nR}{2} \times 40 = 20nR \]

### Final Answer
Thus, the work done by the gas as the temperature changes from \(7^\circ C\) to \(47^\circ C\) is:
\[ W = 20nR \]