In certain region of space there are n number of molecules per unit volume. The temperature of the gas is T. The pressure of the gas will be
(k = Boltzmann's constant and R = universal gas constant)

Options

A. nRT
B. nKT
C. `(nT)/(K)`
D. `(nT)/(R)`

Answer

To solve the problem, we need to derive the expression for the pressure of a gas given the number of molecules per unit volume (n) and the temperature (T). We will use the ideal gas law and the relationship between the universal gas constant (R) and Boltzmann's constant (k).

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = n_gRT \]
where \(P\) is the pressure, \(V\) is the volume, \(n_g\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature.

2. **Relate Moles to Number of Molecules**:
The number of moles \(n_g\) can be expressed in terms of the number of molecules \(N\) using Avogadro's number \(N_A\):
\[ n_g = \frac{N}{N_A} \]
where \(N\) is the total number of molecules in the volume \(V\).

3. **Substituting into the Ideal Gas Law**:
Substituting \(n_g\) into the ideal gas law gives:
\[ PV = \left(\frac{N}{N_A}\right)RT \]

4. **Rearranging the Equation**:
Rearranging this equation to solve for pressure \(P\):
\[ P = \frac{NRT}{VN_A} \]

5. **Using the Definition of Boltzmann's Constant**:
We know that Boltzmann's constant \(k\) is related to \(R\) by:
\[ k = \frac{R}{N_A} \]
Thus, we can substitute \(R\) in terms of \(k\):
\[ P = \frac{NkT}{V} \]

6. **Considering Volume per Unit Volume**:
If we consider the volume \(V = 1\) (as given in the problem), we can simplify the equation:
\[ P = NkT \]

7. **Final Expression for Pressure**:
Therefore, the pressure of the gas in terms of the number of molecules per unit volume \(n\) and temperature \(T\) is:
\[ P = nkT \]

### Conclusion:
The pressure of the gas is given by the formula:
\[ P = nkT \]