For an ideal gas the slope of V-T graph during an adiabatic process is (dV)/(dT) = -m` at a point where volume and temperature are `V_(0)` and `T_(0)`. Find the value of `C_(p)` of the gas. It is given that m is a positive number.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Adiabatic Process for an Ideal Gas
In an adiabatic process, there is no heat exchange with the surroundings. For an ideal gas, the relationship between volume (V), temperature (T), and pressure (P) can be described using the adiabatic condition.

### Step 2: Use the Adiabatic Condition
The adiabatic condition for an ideal gas can be expressed as:
\[ PV^{\gamma} = \text{constant} \]
where \(\gamma = \frac{C_p}{C_v}\) is the heat capacity ratio.

### Step 3: Differentiate the Adiabatic Condition
To find the slope of the V-T graph, we need to express the relationship between volume and temperature. We can differentiate the equation \(PV^{\gamma} = \text{constant}\) with respect to T.

Using the ideal gas law \(PV = nRT\), we can express pressure \(P\) in terms of \(V\) and \(T\):
\[ P = \frac{nRT}{V} \]
Substituting this into the adiabatic condition gives:
\[ \frac{nRT}{V} V^{\gamma} = \text{constant} \]

### Step 4: Find the Slope
We are given that the slope of the V-T graph is:
\[ \frac{dV}{dT} = -m \]
At the point where \(V = V_0\) and \(T = T_0\), we can express this as:
\[ \frac{dV}{dT} = \frac{dV}{dT} \bigg|_{(V_0, T_0)} = -m \]

### Step 5: Relate the Slope to Heat Capacities
From thermodynamic relations, we know:
\[ \frac{dV}{dT} = -\frac{V_0}{T_0}(\gamma - 1) \]
Setting this equal to \(-m\):
\[ -m = -\frac{V_0}{T_0}(\gamma - 1) \]
Thus, we have:
\[ m = \frac{V_0}{T_0}(\gamma - 1) \]

### Step 6: Solve for \(C_p\)
We know that:
\[ \gamma = \frac{C_p}{C_v} \]
And since \(C_v = C_p - R\), we can express \(\gamma\) as:
\[ \gamma = \frac{C_p}{C_p - R} \]
Substituting this into our equation gives:
\[ m = \frac{V_0}{T_0}\left(\frac{C_p}{C_p - R} - 1\right) \]
Simplifying this leads to:
\[ m = \frac{V_0}{T_0}\left(\frac{R}{C_p - R}\right) \]
Rearranging gives:
\[ C_p = \frac{m T_0}{V_0} R + R \]

### Final Expression for \(C_p\)
Thus, the final expression for \(C_p\) is:
\[ C_p = \frac{m T_0}{V_0} R + R \]