One mole of an ideal monatomic gas at temperature `T_0` expands slowly according to the law `P = kV` (k is constant). If the final temperature is `4T_0` then heat supplied to gas is

Options

A. `2RT_(0)`
B. `(3)/(2)RT_(0)`
C. `6RT_(0)`
D. `(RT_(0))/(2)`

Answer

To solve the problem, we need to find the heat supplied to one mole of an ideal monatomic gas that expands according to the law \( P = kV \) and reaches a final temperature of \( 4T_0 \). We will follow these steps:

### Step 1: Identify the Polytropic Process
The given relationship \( P = kV \) indicates a polytropic process. We can express this in terms of the polytropic index \( n \). The general form of a polytropic process is given by:

\[ PV^n = \text{constant} \]

From the equation \( P = kV \), we can rearrange it to show that:

\[ P = kV \implies PV = kV^2 \]

This suggests that the polytropic index \( n = -1 \) (since \( P \) is inversely proportional to \( V \)).

### Step 2: Calculate Molar Specific Heat
For a polytropic process, the molar specific heat \( C \) can be calculated using the formula:

\[ C = C_v + \frac{R}{1 - n} \]

Where:
- \( C_v \) is the molar specific heat at constant volume.
- \( R \) is the gas constant.
- \( n \) is the polytropic index.

For a monatomic ideal gas, the molar specific heat at constant volume \( C_v \) is:

\[ C_v = \frac{3}{2}R \]

Substituting \( n = -1 \):

\[ C = C_v + \frac{R}{1 - (-1)} = C_v + \frac{R}{2} = \frac{3}{2}R + \frac{R}{2} = 2R \]

### Step 3: Calculate the Change in Temperature
The change in temperature \( \Delta T \) is given by:

\[ \Delta T = T_f - T_i = 4T_0 - T_0 = 3T_0 \]

### Step 4: Calculate the Heat Supplied
The heat supplied \( Q \) during the process can be calculated using the formula:

\[ Q = nC\Delta T \]

Where:
- \( n \) is the number of moles (which is 1 in this case).
- \( C \) is the molar specific heat calculated in Step 2.
- \( \Delta T \) is the change in temperature calculated in Step 3.

Substituting the values:

\[ Q = 1 \cdot (2R) \cdot (3T_0) = 6RT_0 \]

### Final Answer
Thus, the heat supplied to the gas is:

\[ Q = 6RT_0 \]