If `p=(RT)/(V-b)e^(-alphaV//RT)` , then dimensional formula of `alpha` is same as that of

Options

A. `P`
B. `R`
C. `T`
D. `V`

Answer

To find the dimensional formula of \( \alpha \) in the equation

\[ p = \frac{RT}{(V-b)e^{-\frac{\alpha V}{RT}}} \]

we need to analyze the equation step by step.

### Step 1: Understand the equation
The equation given is for pressure \( p \). In this equation, \( R \) is the universal gas constant, \( T \) is the temperature, \( V \) is the volume, and \( b \) is a constant. The term \( e^{-\frac{\alpha V}{RT}} \) must be dimensionless, as the argument of an exponential function must be unitless.

### Step 2: Set the argument of the exponential to be dimensionless
The argument of the exponential function is

\[ -\frac{\alpha V}{RT} \]

For this to be dimensionless, the dimensions of \( \alpha V \) must equal the dimensions of \( RT \).

### Step 3: Write the dimensions
Let’s denote the dimensions of the quantities involved:
- The dimension of pressure \( P \) is \( [M L^{-1} T^{-2}] \).
- The dimension of volume \( V \) is \( [L^3] \).
- The dimension of temperature \( T \) is \( [\Theta] \).
- The dimension of the universal gas constant \( R \) can be derived from the ideal gas law \( PV = nRT \). Thus, the dimension of \( R \) is

\[ [R] = \frac{[P][V]}{[n][T]} = \frac{[M L^{-1} T^{-2}][L^3]}{[N][\Theta]} = [M L^{2} T^{-2} N^{-1} \Theta^{-1}] \]

### Step 4: Set up the equation for dimensions
From the requirement that \( \alpha V \) must have the same dimensions as \( RT \), we can write:

\[ [\alpha V] = [RT] \]

Substituting the dimensions we found:

\[ [\alpha][L^3] = [M L^{2} T^{-2} N^{-1} \Theta^{-1}] \]

### Step 5: Solve for the dimensions of \( \alpha \)
Now, we can isolate the dimensions of \( \alpha \):

\[ [\alpha] = \frac{[M L^{2} T^{-2} N^{-1} \Theta^{-1}]}{[L^3]} = [M L^{-1} T^{-2} N^{-1} \Theta^{-1}] \]

### Conclusion
Thus, the dimensional formula of \( \alpha \) is

\[ [M L^{-1} T^{-2} N^{-1} \Theta^{-1}] \]