The value of `gamma = C_(P)//C_(V)` is `4//3` for an adaibatic process of an ideal gas for which internal energy `U = K + nPV`. Find the value of `n` (`K` is constant) :

Explanation

To find the value of \( n \) in the equation for internal energy \( U = K + nPV \) given that \( \gamma = \frac{C_P}{C_V} = \frac{4}{3} \) for an adiabatic process of an ideal gas, we can follow these steps:

### Step 1: Understand the relationship for an adiabatic process
For an adiabatic process, we have the relation:
\[ PV^\gamma = \text{constant} \]
Given that \( \gamma = \frac{4}{3} \), we can write:
\[ PV^{\frac{4}{3}} = \text{constant} \]

### Step 2: Differentiate the adiabatic relation
Differentiating the equation \( PV^{\frac{4}{3}} = \text{constant} \) gives:
\[ \frac{d(PV^{\frac{4}{3}})}{dt} = 0 \]
Using the product rule, we have:
\[ P \frac{d(V^{\frac{4}{3}})}{dt} + V^{\frac{4}{3}} \frac{dP}{dt} = 0 \]
This can be rewritten as:
\[ P \cdot \frac{4}{3} V^{\frac{1}{3}} \frac{dV}{dt} + V^{\frac{4}{3}} \frac{dP}{dt} = 0 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ \frac{4}{3} P V^{\frac{1}{3}} \frac{dV}{dt} = -V^{\frac{4}{3}} \frac{dP}{dt} \]
Dividing both sides by \( PV^{\frac{4}{3}} \):
\[ \frac{4}{3} \frac{1}{V} \frac{dV}{dt} = -\frac{1}{P} \frac{dP}{dt} \]

### Step 4: Relate internal energy change to work done
For an adiabatic process, the first law of thermodynamics states:
\[ \Delta Q = 0 \implies \Delta U = -\Delta W \]
The change in internal energy \( \Delta U \) can be expressed as:
\[ \Delta U = nR \Delta T \]
And the work done \( \Delta W \) by the gas is given by:
\[ \Delta W = P \Delta V \]

### Step 5: Substitute the expression for internal energy
Given \( U = K + nPV \), we differentiate:
\[ \Delta U = nP \Delta V + nV \Delta P \]
Setting this equal to the work done:
\[ nP \Delta V + nV \Delta P = -P \Delta V \]

### Step 6: Solve for \( n \)
Rearranging gives:
\[ nP \Delta V + nV \Delta P + P \Delta V = 0 \]
Factoring out \( \Delta V \):
\[ (n + 1)P \Delta V + nV \Delta P = 0 \]
For this to hold for all \( \Delta V \) and \( \Delta P \), we can set the coefficients to zero. Thus:
1. \( n + 1 = 0 \) leads to \( n = -1 \) (not valid for our case).
2. \( nV = -P \) leads to \( n = 3 \).

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{3} \]