The relation between internal energy `U`, pressure `P` and volume `V` of a gas in an adiabatic processes is : `U = a+ bPV` where a = b = `3` . Calculate the greatest integer of the ratio of specific heats `[gamma]` .

Explanation

To solve the problem, we start with the given relation between internal energy \( U \), pressure \( P \), and volume \( V \) of the gas in an adiabatic process:

\[ U = a + bPV \]

where \( a = 3 \) and \( b = 3 \).

### Step 1: Write down the expression for internal energy.
Given:
\[ U = 3 + 3PV \]

### Step 2: Use the first law of thermodynamics for an adiabatic process.
For an adiabatic process, the heat exchange \( dq = 0 \). According to the first law of thermodynamics:
\[ dq = dU + PdV = 0 \]
This implies:
\[ dU + PdV = 0 \quad \Rightarrow \quad dU = -PdV \]

### Step 3: Differentiate the expression for internal energy.
Differentiating \( U \):
\[ dU = \frac{dU}{dP}dP + \frac{dU}{dV}dV \]
From our expression for \( U \):
\[ dU = 3 \cdot d(PV) = 3(PdV + VdP) \]
Thus:
\[ dU = 3PdV + 3VdP \]

### Step 4: Set the expressions for \( dU \) equal.
From the first law:
\[ 3PdV + 3VdP = -PdV \]
Rearranging gives:
\[ 3PdV + PdV + 3VdP = 0 \]
\[ (3 + 1)PdV + 3VdP = 0 \]
\[ 4PdV + 3VdP = 0 \]

### Step 5: Rearranging the equation.
Dividing through by \( PV \):
\[ \frac{4}{V}dV + \frac{3}{P}dP = 0 \]

### Step 6: Integrate the equation.
This can be rearranged to:
\[ \frac{dV}{V} = -\frac{3}{4}\frac{dP}{P} \]
Integrating both sides gives:
\[ \ln V = -\frac{3}{4} \ln P + C \]
Exponentiating:
\[ V^{\frac{4}{3}} P^{3} = K \]
where \( K \) is a constant.

### Step 7: Relate to the specific heats.
For an ideal gas, the relationship between specific heats \( \gamma \) is given by:
\[ \gamma = \frac{C_p}{C_v} \]
From the derived equation, we have:
\[ PV^{\frac{4}{3}} = \text{constant} \]
This indicates that:
\[ \gamma = \frac{b + 1}{b} = \frac{4}{3} \]

### Step 8: Calculate the greatest integer of \( \gamma \).
Calculating \( \gamma \):
\[ \gamma = \frac{4}{3} \approx 1.33 \]
Thus, the greatest integer of \( \gamma \) is:
\[ \lfloor \gamma \rfloor = 1 \]

### Final Answer:
The greatest integer of the ratio of specific heats \( \gamma \) is \( 1 \).

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