The relation between U, p and V for an ideal gas in an adiabatic process is given by relation `U=a+bpV`. Find the value of adiabatic exponent `(gamma)` of this gas.

Options

A. `(b+1)/(b)`
B. `(b+1)/(a)`
C. `(a+1)/(b)`
D. `(a)/(a+b)`

Answer

To find the value of the adiabatic exponent \((\gamma)\) for an ideal gas in an adiabatic process given the relation \(U = a + bPV\), we can follow these steps:

### Step 1: Understand the relationship
The internal energy \(U\) of an ideal gas in an adiabatic process is given by the equation:
\[ U = a + bPV \]
where \(a\) and \(b\) are constants, \(P\) is the pressure, and \(V\) is the volume.

### Step 2: Recall the first law of thermodynamics
In an adiabatic process, the heat exchange \(dQ\) is zero. Therefore, the first law of thermodynamics states:
\[ dU = dQ - dW \]
Since \(dQ = 0\), we have:
\[ dU = -dW \]

### Step 3: Express work done in terms of \(PV\)
The work done \(dW\) in an adiabatic process can be expressed as:
\[ dW = PdV + VdP \]
For an ideal gas, we can also relate \(PV\) to temperature and the number of moles \(n\) using the ideal gas law:
\[ PV = nRT \]

### Step 4: Relate \(dU\) to \(dT\)
From the ideal gas law, we can express the change in internal energy \(dU\) as:
\[ dU = nC_v dT \]
where \(C_v\) is the molar heat capacity at constant volume.

### Step 5: Use the adiabatic condition
For an adiabatic process, we know that:
\[ PV^{\gamma} = \text{constant} \]
This implies that:
\[ d(PV^{\gamma}) = 0 \]
From this, we can derive relationships between \(P\), \(V\), and \(T\).

### Step 6: Compare coefficients
By comparing the expressions derived from the internal energy equation \(U = a + bPV\) with the derived expressions involving \(C_v\) and the adiabatic condition, we can find the relationship between \(b\) and \(\gamma\).

### Step 7: Solve for \(\gamma\)
From the relationship derived, we find:
\[ \frac{1}{b} = \gamma - 1 \]
Thus, rearranging gives:
\[ \gamma = 1 + \frac{1}{b} \]

### Final Result
The value of the adiabatic exponent \(\gamma\) is:
\[ \gamma = 1 + \frac{1}{b} \]