Gas constant 'R' has dimensions :

Options

A. `M^(1)L^(2)T^(2)K^(-1)`
B. `M^(1)L^(2)T^(-2)K^(-1)`
C. `M^(-1)L^(-2)T^(-2)K^(-1)`
D. `M^(1)L^(1)T^(2)K^(1)`

Answer

To find the dimensions of the gas constant 'R', we start with the ideal gas law equation:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles (dimensionless)
- \( R \) = Gas constant
- \( T \) = Temperature

### Step 1: Identify the dimensions of each term in the equation.

1. **Pressure (P)**: Pressure is defined as force per unit area.
- Force (F) has dimensions of mass (M) times acceleration (LT^-2), so:
\[ [F] = MLT^{-2} \]
- Area (A) has dimensions of length squared (L^2), so:
\[ [A] = L^2 \]
- Therefore, the dimension of pressure (P) is:
\[ [P] = \frac{[F]}{[A]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

2. **Volume (V)**: Volume has dimensions of length cubed (L^3).
\[ [V] = L^3 \]

### Step 2: Combine the dimensions of pressure and volume.

Using the dimensions of pressure and volume, we can find the dimensions of \( PV \):
\[ [PV] = [P][V] = (ML^{-1}T^{-2})(L^3) = ML^{2}T^{-2} \]

### Step 3: Identify the dimensions of temperature (T).

The dimension of temperature (T) is given as:
\[ [T] = K \] (Kelvin)

### Step 4: Rearrange the ideal gas law to solve for R.

From the ideal gas law, we can express R as:
\[ R = \frac{PV}{nT} \]

Since \( n \) is dimensionless, we can ignore it in our dimensional analysis.

### Step 5: Substitute the dimensions into the equation.

Now we can substitute the dimensions we found into the equation for R:
\[ [R] = \frac{[PV]}{[T]} = \frac{ML^{2}T^{-2}}{K} \]

### Step 6: Write the final expression for the dimensions of R.

Thus, the dimensions of the gas constant R are:
\[ [R] = ML^{2}T^{-2}K^{-1} \]

### Conclusion

The final dimensions of the gas constant R are:
\[ [R] = M^1 L^2 T^{-2} K^{-1} \]