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 step by step, we will analyze the adiabatic process of the gas and derive the expression for the molar specific heat at constant pressure \( C_p \).

### Step 1: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings. The relationship between pressure \( P \), volume \( V \), and temperature \( T \) for an ideal gas undergoing an adiabatic process is given by:
\[ PV^\gamma = \text{constant} \]
where \( \gamma = \frac{C_p}{C_v} \).

### Step 2: Use the Ideal Gas Law
From the ideal gas law, we have:
\[ PV = nRT \]
We can express \( P \) in terms of \( V \) and \( T \):
\[ P = \frac{nRT}{V} \]

### Step 3: Substitute into the Adiabatic Equation
Substituting \( P \) into the adiabatic equation gives:
\[ \left(\frac{nRT}{V}\right)V^\gamma = C \]
Simplifying this, we get:
\[ nRTV^{\gamma - 1} = C \]

### Step 4: Rearranging the Equation
Rearranging the equation, we find:
\[ TV^{\gamma - 1} = \frac{C}{nR} = C' \]
This shows the relationship between \( T \) and \( V \) during the adiabatic process.

### Step 5: Differentiate the Equation
To find the slope of the \( V-T \) graph, we differentiate the equation \( TV^{\gamma - 1} = C' \) with respect to \( T \):
\[ \frac{d}{dT}(TV^{\gamma - 1}) = 0 \]
Using the product rule:
\[ V^{\gamma - 1} + T(\gamma - 1)V^{\gamma - 2}\frac{dV}{dT} = 0 \]

### Step 6: Solve for \(\frac{dV}{dT}\)
Rearranging gives:
\[ \frac{dV}{dT} = -\frac{V^{\gamma - 1}}{T(\gamma - 1)V^{\gamma - 2}} = -\frac{V}{T(\gamma - 1)} \]
Thus, the slope of the \( V-T \) graph is:
\[ \left| \frac{dV}{dT} \right| = \frac{V}{T(\gamma - 1)} \]

### Step 7: Substitute Known Values
At the point where \( V = V_0 \) and \( T = T_0 \), we have:
\[ m = \frac{V_0}{T_0(\gamma - 1)} \]

### Step 8: Express \(\gamma\) in Terms of \(C_p\) and \(C_v\)
Recall that:
\[ \gamma = \frac{C_p}{C_v} \]
and
\[ C_p - C_v = R \]
Thus, we can express \( \gamma - 1 \) as:
\[ \gamma - 1 = \frac{C_p - R}{C_v} \]

### Step 9: Solve for \(C_p\)
Substituting this back into our equation for \( m \):
\[ m = \frac{V_0}{T_0 \left(\frac{C_p - R}{C_v}\right)} \]
Rearranging gives:
\[ C_p = \frac{R V_0}{T_0 m} + R \]

### Step 10: Final Expression for \(C_p\)
Thus, we can express \( C_p \) as:
\[ C_p = \frac{R (V_0 + m T_0)}{V_0} \]

### Conclusion
The molar specific heat at constant pressure \( C_p \) is given by:
\[ C_p = R \frac{V_0 + m T_0}{V_0} \]