One mole of an ideal gas (mono-atomic) at temperature `T_0` expands slowly according to law `P^2 = c T` (c is constant). If final temperature is `2T_0` heat supplied to gas is:

Options

A. `2 RT_0`
B. `3/2 RT_0`
C. `RT_0`
D. `(RT_0)/2`

Explanation

To solve the problem, we need to find the heat supplied to one mole of an ideal monoatomic gas that expands according to the law \( P^2 = cT \) from an initial temperature \( T_0 \) to a final temperature \( 2T_0 \).

### Step-by-step Solution:

1. **Identify the Given Information:**
- Initial temperature, \( T_1 = T_0 \)
- Final temperature, \( T_2 = 2T_0 \)
- Number of moles, \( n = 1 \)
- The relationship given is \( P^2 = cT \).

2. **Determine the Polytropic Index \( x \):**
- From the ideal gas law, we know \( PV = nRT \).
- Rearranging gives \( T = \frac{PV}{R} \).
- Substitute \( T \) into the given equation:
\[ P^2 = c \left(\frac{PV}{R}\right) \]
- This simplifies to:
\[ P^2 = \frac{cPV}{R} \]
- Rearranging gives:
\[ P^{1} V^{-1} = \frac{c}{R} \]
- This indicates that the polytropic index \( x = -1 \).

3. **Calculate the Polytropic Heat Capacity \( C \):**
- The formula for polytropic heat capacity is:
\[ C = C_v + \frac{R}{1 - x} \]
- For a monoatomic ideal gas, \( C_v = \frac{3R}{2} \).
- Substitute \( x = -1 \):
\[ C = \frac{3R}{2} + \frac{R}{1 - (-1)} = \frac{3R}{2} + \frac{R}{2} = \frac{4R}{2} = 2R \]

4. **Calculate the Change in Temperature \( \Delta T \):**
- The change in temperature is:
\[ \Delta T = T_2 - T_1 = 2T_0 - T_0 = T_0 \]

5. **Calculate the Heat Supplied \( Q \):**
- The heat supplied to the gas can be calculated using:
\[ Q = nC\Delta T \]
- Substituting the values:
\[ Q = 1 \cdot (2R) \cdot (T_0) = 2RT_0 \]

### Final Result:
The heat supplied to the gas is \( Q = 2RT_0 \).

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