An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation `p=kv`. Show that the molar heat capacity of the gas for the process is given by `(C=C_v + ( R)/(2))`.

Explanation

To solve the problem, we need to derive the expression for the molar heat capacity \( C \) of an ideal gas undergoing a process defined by the equation \( p = kV \). Here’s a step-by-step solution:

### Step 1: Start with the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)), this simplifies to:
\[ PV = RT \]

### Step 2: Substitute the given relationship \( p = kV \)
From the problem, we have:
\[ p = kV \]
Substituting this into the ideal gas law gives:
\[ (kV)V = RT \]
This simplifies to:
\[ kV^2 = RT \]

### Step 3: Differentiate with respect to \( T \)
We differentiate both sides of the equation \( kV^2 = RT \) with respect to \( T \):
\[ \frac{d}{dT}(kV^2) = \frac{d}{dT}(RT) \]
Using the product rule on the left side, we get:
\[ k \cdot 2V \frac{dV}{dT} = R \]
This can be rearranged to find \( \frac{dV}{dT} \):
\[ \frac{dV}{dT} = \frac{R}{2kV} \]

### Step 4: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ dQ = dU + dW \]
For one mole of gas, the change in internal energy \( dU \) is given by:
\[ dU = C_v dT \]
The work done \( dW \) during an expansion is:
\[ dW = pdV \]
Substituting \( p = kV \) gives:
\[ dW = kV dV \]

### Step 5: Substitute \( dW \) and \( dU \) into the First Law
Now we can write:
\[ dQ = C_v dT + kV dV \]
Substituting \( dV \) from Step 3:
\[ dQ = C_v dT + kV \left(\frac{R}{2kV}\right) dT \]
This simplifies to:
\[ dQ = C_v dT + \frac{R}{2} dT \]

### Step 6: Factor out \( dT \) to find \( C \)
Factoring \( dT \) out gives:
\[ dQ = \left(C_v + \frac{R}{2}\right) dT \]
Thus, the molar heat capacity \( C \) for the process is:
\[ C = C_v + \frac{R}{2} \]

### Conclusion
We have shown that the molar heat capacity of the gas for the process is given by:
\[ C = C_v + \frac{R}{2} \]