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 given relation
The internal energy \( U \) is expressed as:
\[ U = a + bPV \]
where \( a \) is a constant, \( b \) is a coefficient, \( P \) is the pressure, and \( V \) is the volume.

### Step 2: Recall the relationship for internal energy in an adiabatic process
For an ideal gas, the change in internal energy \( \Delta U \) can be expressed in terms of temperature:
\[ \Delta U = nC_V \Delta T \]
where \( C_V \) is the molar heat capacity at constant volume, \( n \) is the number of moles, and \( \Delta T \) is the change in temperature.

### Step 3: Relate \( PV \) to temperature
Using the ideal gas law, we know:
\[ PV = nRT \]
where \( R \) is the universal gas constant.

### Step 4: Substitute \( PV \) in the internal energy equation
From the ideal gas law, we can express \( T \) as:
\[ T = \frac{PV}{nR} \]
Substituting this into the expression for internal energy gives:
\[ U = a + bPV = a + b \left( nR \frac{T}{n} \right) = a + b \frac{PV}{n} \]

### Step 5: Express \( U \) in terms of \( PV \)
We can also express the internal energy in terms of \( PV \):
\[ U = \frac{PV}{\gamma - 1} \]
This comes from the relation for an ideal gas in an adiabatic process, where \( PV^{\gamma} = \text{constant} \).

### Step 6: Compare the two expressions for internal energy
Now we have two expressions for \( U \):
1. \( U = a + bPV \)
2. \( U = \frac{PV}{\gamma - 1} \)

Setting these equal gives:
\[ a + bPV = \frac{PV}{\gamma - 1} \]

### Step 7: Rearranging the equation
Rearranging this equation, we can isolate terms involving \( PV \):
\[ bPV = \frac{PV}{\gamma - 1} - a \]
Assuming \( PV \neq 0 \), we can divide through by \( PV \):
\[ b = \frac{1}{\gamma - 1} - \frac{a}{PV} \]

### Step 8: Solve for \( \gamma \)
From the expression \( b = \frac{1}{\gamma - 1} \), we can rearrange to find \( \gamma \):
\[ b(\gamma - 1) = 1 \]
\[ b\gamma - b = 1 \]
\[ b\gamma = 1 + b \]
\[ \gamma = \frac{1 + b}{b} \]

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