A gas is undergoing an adiabatic process. At a certain stage, the volume and absolute temperature of the gas are `V_(0), T_(0)` and the magnitude of the slope of the V-T curve is m. molar specific heat of the gas at constant pressure is [Assume the volume of the gas is taken on the y-axis and absolute temperature of the gas taken on x-axis]

Options

A. `(mRT_(0))/(V_(0))`
B. `(MRT_(0))/( 2V_(0))`
C. `((V_(0)+mT_(0))R)/(V_(0))`
D. `((V_(0)+mT_(0))R)/(2V_(0))`

Answer

To solve the problem, we need to find the molar specific heat of the gas at constant pressure (C_p) during an adiabatic process, given the volume (V_0), absolute temperature (T_0), and the slope (m) of the V-T curve.

### Step-by-Step Solution:

1. **Understanding the Adiabatic Process**:
- In an adiabatic process, there is no heat exchange with the surroundings, which means \(Q = 0\).
- According to the first law of thermodynamics, we have:
\[ dQ = dU + dW \]
Since \(dQ = 0\), it follows that:
\[ dU + dW = 0 \quad \Rightarrow \quad dU = -dW \]

2. **Expressing Work Done (dW)**:
- The work done in a process can be expressed as:
\[ dW = P dV \]
- Thus, we can rewrite the equation as:
\[ dU = -P dV \]

3. **Differentiating Internal Energy (U)**:
- The change in internal energy \(dU\) can be expressed in terms of temperature:
\[ dU = C_V dT \]
- Therefore, we have:
\[ C_V dT = -P dV \]

4. **Using the Slope of the V-T Curve**:
- The slope of the V-T curve is given as \(m\), which is defined as:
\[ m = \frac{dV}{dT} \]
- Rearranging gives:
\[ dV = m dT \]

5. **Substituting dV in the Internal Energy Equation**:
- Substituting \(dV\) into the equation for \(dU\):
\[ C_V dT = -P (m dT) \]
- Dividing both sides by \(dT\) (assuming \(dT \neq 0\)):
\[ C_V = -P m \]

6. **Using the Ideal Gas Law**:
- From the ideal gas law, we know:
\[ PV = nRT \]
- Rearranging gives:
\[ P = \frac{nRT}{V} \]
- Substituting this into the equation for \(C_V\):
\[ C_V = -\left(\frac{nRT}{V}\right) m \]

7. **Finding the Molar Specific Heat at Constant Pressure (C_p)**:
- The relationship between \(C_p\) and \(C_V\) is given by:
\[ C_p = C_V + R \]
- Substituting \(C_V\) into the equation:
\[ C_p = -\left(\frac{nRT}{V}\right) m + R \]

8. **Final Expression**:
- We can express \(C_p\) in terms of \(T_0\) and \(V_0\):
\[ C_p = -\left(\frac{nRT_0}{V_0}\right) m + R \]
- Rearranging gives:
\[ C_p = m \frac{T_0}{V_0} + R \]

### Final Result:
Thus, the molar specific heat of the gas at constant pressure is:
\[ C_p = m \frac{T_0}{V_0} + R \]