The bulk modulus of a gas is defined as B = -VdP/dV. For an adiabatic process the variation of B is proportional to `P^(n)`. For an idea gas, n is

Options

A. 0
B. 1
C. `5/2`
D. `2`

Explanation

To solve the problem, we start with the definition of the bulk modulus \( B \) for a gas:

\[ B = -V \frac{dP}{dV} \]

For an adiabatic process, we know that the volume \( V \) is proportional to the pressure \( P \) raised to the power \( n \). This can be expressed mathematically as:

\[ V \propto P^n \quad \text{or} \quad V = k P^n \]

where \( k \) is a constant.

### Step 1: Use the adiabatic condition

For an adiabatic process, we have the relation:

\[ PV^\gamma = \text{constant} \]

where \( \gamma \) is the heat capacity ratio \( C_p/C_v \).

### Step 2: Differentiate the adiabatic condition

We differentiate the equation \( PV^\gamma = \text{constant} \) with respect to \( V \):

\[ \frac{d}{dV}(PV^\gamma) = 0 \]

Using the product rule, we get:

\[ P \frac{d(V^\gamma)}{dV} + V^\gamma \frac{dP}{dV} = 0 \]

Calculating \( \frac{d(V^\gamma)}{dV} \):

\[ \frac{d(V^\gamma)}{dV} = \gamma V^{\gamma-1} \]

Substituting this back into our differentiated equation gives:

\[ P \cdot \gamma V^{\gamma-1} + V^\gamma \frac{dP}{dV} = 0 \]

### Step 3: Rearranging the equation

Rearranging the equation yields:

\[ V^\gamma \frac{dP}{dV} = -P \cdot \gamma V^{\gamma-1} \]

Dividing both sides by \( V^\gamma \):

\[ \frac{dP}{dV} = -\frac{\gamma P}{V} \]

### Step 4: Substitute into the bulk modulus equation

Now we substitute \( \frac{dP}{dV} \) into the bulk modulus equation:

\[ B = -V \left(-\frac{\gamma P}{V}\right) = \gamma P \]

### Step 5: Relate bulk modulus to pressure

From our derivation, we see that:

\[ B \propto P \]

This indicates that the bulk modulus \( B \) is directly proportional to \( P^1 \).

### Step 6: Conclusion

Thus, we can conclude that for an ideal gas undergoing an adiabatic process, the value of \( n \) is:

\[ \boxed{1} \]