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 `T_(0)` to `4T_(0)` is

Options

A. work done by the gas is `3nRT_(0)`
B. heat supplied to the gas is `6nRT_(0)`
C. work done by the gas is `(3)/(2)nRT_(0)`
D. heat supplied to the gas is `(3)/(2)nRT_(0)`

Answer

To solve the problem of finding the work done by an ideal gas as it undergoes a process where the temperature changes with volume according to the equation \( T = kV^2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**:
We are given that \( T = kV^2 \). This means that the temperature of the gas is proportional to the square of its volume.

2. **Use the Ideal Gas Law**:
The ideal gas law is given by:
\[ PV = nRT \]
Substituting \( T = kV^2 \) into the ideal gas law:
\[ PV = nR(kV^2) \]
Rearranging gives:
\[ P = \frac{nRkV^2}{V} = nRkV \]

3. **Determine the Volumes at Given Temperatures**:
- For \( T = T_0 \):
\[ T_0 = kV_1^2 \implies V_1^2 = \frac{T_0}{k} \implies V_1 = \sqrt{\frac{T_0}{k}} = V_0 \]
- For \( T = 4T_0 \):
\[ 4T_0 = kV_2^2 \implies V_2^2 = \frac{4T_0}{k} \implies V_2 = \sqrt{\frac{4T_0}{k}} = 2\sqrt{\frac{T_0}{k}} = 2V_0 \]

4. **Calculate the Work Done**:
The work done \( W \) by the gas during this process can be expressed as:
\[ W = \int_{V_1}^{V_2} P \, dV \]
Substituting \( P = nRkV \):
\[ W = \int_{V_0}^{2V_0} nRkV \, dV \]
This integral can be computed as follows:
\[ W = nRk \int_{V_0}^{2V_0} V \, dV = nRk \left[ \frac{V^2}{2} \right]_{V_0}^{2V_0} \]
Evaluating the limits:
\[ W = nRk \left( \frac{(2V_0)^2}{2} - \frac{(V_0)^2}{2} \right) = nRk \left( \frac{4V_0^2}{2} - \frac{V_0^2}{2} \right) = nRk \left( \frac{3V_0^2}{2} \right) \]

5. **Substitute \( V_0 \)**:
Recall that \( V_0 = \sqrt{\frac{T_0}{k}} \):
\[ V_0^2 = \frac{T_0}{k} \]
Thus,
\[ W = nRk \left( \frac{3}{2} \cdot \frac{T_0}{k} \right) = \frac{3}{2} nRT_0 \]

### Final Answer:
The work done by the gas as the temperature changes from \( T_0 \) to \( 4T_0 \) is:
\[ W = \frac{3}{2} nRT_0 \]