The viscosity `eta` of a gas depends on the long - range attractive part of the intermolecular force, which varies with molecular separation r according to `F=mur^(-n)` where n is a number and `mu` is a constant. If `eta` is a function of the mass m of the molecules, their mean speed v, and the constant `mu` than which of the following is correct-

Options

A. `eta prop m^(n+1)v^(n+3)mu^(n-2)`
B. `eta prop m^((n+1)/(n-1))v^((n+3)/(n-1))mu^((-2)/(n-1))`
C. `eta prop m^(n)v^(-n)mu^(-2)`
D. `eta prop mv mu^(-n)`

Answer

To solve the problem regarding the viscosity \( \eta \) of a gas and its dependence on mass \( m \), mean speed \( v \), and a constant \( \mu \), we will perform dimensional analysis. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship
The viscosity \( \eta \) is said to depend on the mass \( m \) of the molecules, their mean speed \( v \), and a constant \( \mu \). We can express this relationship as:
\[ \eta = k \cdot m^a \cdot v^b \cdot \mu^c \]
where \( k \) is a proportionality constant and \( a, b, c \) are the powers we need to determine.

### Step 2: Determine the dimensions of viscosity
Viscosity \( \eta \) can be defined through the formula for viscous force:
\[ F = 6 \pi \eta r v \]
From this, we can rearrange to find \( \eta \):
\[ \eta = \frac{F}{6 \pi r v} \]
The dimensions of force \( F \) are:
\[ [F] = M L T^{-2} \]
The dimensions of radius \( r \) are:
\[ [r] = L \]
The dimensions of speed \( v \) are:
\[ [v] = L T^{-1} \]
Substituting these into the viscosity formula gives:
\[ [\eta] = \frac{M L T^{-2}}{L \cdot L T^{-1}} = \frac{M L T^{-2}}{L^2 T^{-1}} = M L^{-1} T^{-1} \]

### Step 3: Write down the dimensions of the variables
Now we have:
- \( [m] = M \)
- \( [v] = L T^{-1} \)
- \( [\mu] \) needs to be determined from the force equation \( F = \mu r^{-n} \).

### Step 4: Determine the dimensions of \( \mu \)
From the equation \( F = \mu r^{-n} \), we can express \( \mu \) as:
\[ \mu = \frac{F}{r^{-n}} = F \cdot r^n \]
Thus, the dimensions of \( \mu \) are:
\[ [\mu] = [F] \cdot [r]^n = (M L T^{-2}) \cdot (L)^n = M L^{n+1} T^{-2} \]

### Step 5: Substitute dimensions into the equation
Now substituting the dimensions into the equation for \( \eta \):
\[ [M L^{-1} T^{-1}] = [M]^a \cdot [L T^{-1}]^b \cdot [M L^{n+1} T^{-2}]^c \]
This expands to:
\[ M^1 L^{-1} T^{-1} = M^a \cdot L^{b + c(n + 1)} \cdot T^{-b - 2c} \]

### Step 6: Equate the powers of dimensions
Equating the powers of \( M \), \( L \), and \( T \):
1. For \( M \): \( 1 = a + c \)
2. For \( L \): \( -1 = b + c(n + 1) \)
3. For \( T \): \( -1 = -b - 2c \)

### Step 7: Solve the equations
From the first equation:
\[ c = 1 - a \]
Substituting \( c \) into the second and third equations:
- Second equation:
\[ -1 = b + (1 - a)(n + 1) \implies -1 = b + n + 1 - a(n + 1) \implies b = -n - a(n + 1) \]
- Third equation:
\[ -1 = -b - 2(1 - a) \implies -1 = -b - 2 + 2a \implies b = 1 + 2a \]

### Step 8: Substitute and solve for \( a \)
Setting the two expressions for \( b \) equal:
\[ -n - a(n + 1) = 1 + 2a \]
Rearranging gives:
\[ -n - 1 = a(n + 1 + 2) \implies a = \frac{-n - 1}{n + 3} \]

### Step 9: Find \( b \) and \( c \)
Substituting \( a \) back into the equations will yield \( b \) and \( c \).

### Final Result
After solving for \( a, b, c \), we can express \( \eta \) in terms of \( m, v, \mu \) as:
\[ \eta \propto m^{\frac{-n-1}{n+3}} v^{\frac{n+3}{n+1}} \mu^{\frac{-2}{n+1}} \]

### Conclusion
The correct option will be the one that matches this derived relationship.