One mole of an ideal monoatomic gas at temperature `T_0` expands slowly according to the law `p/V` = constant. If the final temperature is `2T_0`, heat supplied to the gas is

Options

A. `2RT_(0)`
B. `RT_(o)`
C. `3/2RT_(0)`
D. `1/2RT_(0)`

Explanation

To solve the problem of finding the heat supplied to one mole of an ideal monoatomic gas that expands according to the law \( \frac{p}{V} = \text{constant} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Conditions**:
- Initial temperature \( T_0 \)
- Final temperature \( T_f = 2T_0 \)
- The process follows the relation \( \frac{p}{V} = \text{constant} \).

2. **Determine the Type of Process**:
- The relation \( \frac{p}{V} = \text{constant} \) suggests that the process is a polytropic process with \( n = -1 \).

3. **Use the Molar Heat Capacity for the Process**:
- For a monoatomic ideal gas, the heat capacity \( C \) during a polytropic process can be expressed as:
\[ C = R \left( \frac{\gamma}{\gamma - 1} + \frac{1}{1 - n} \right) \]
- Here, for a monoatomic gas, \( \gamma = \frac{5}{3} \) and \( n = -1 \).

4. **Calculate the Molar Heat Capacity**:
- Substitute the values into the equation:
\[ C = R \left( \frac{5/3}{5/3 - 1} + \frac{1}{1 - (-1)} \right) \]
- Simplifying gives:
\[ C = R \left( \frac{5/3}{2/3} + \frac{1}{2} \right) = R \left( \frac{5}{2} + \frac{1}{2} \right) = R \cdot 3 = 2R \]

5. **Calculate the Change in Temperature**:
- The change in temperature \( \Delta T \) is:
\[ \Delta T = T_f - T_0 = 2T_0 - T_0 = T_0 \]

6. **Calculate the Heat Supplied**:
- The heat supplied \( \Delta Q \) can be calculated using:
\[ \Delta Q = nC\Delta T \]
- Since \( n = 1 \) (one mole of gas), we have:
\[ \Delta Q = 1 \cdot (2R) \cdot (T_0) = 2RT_0 \]

7. **Final Answer**:
- The heat supplied to the gas is:
\[ \Delta Q = 2RT_0 \]

### Summary:
The heat supplied to the gas during the expansion is \( 2RT_0 \).