We are given the Van der Waals equation of state for a gas:
(P+aV2)(V−b)=RT\left( P + \frac{a}{{V^2}} \right)(V - b) = RT
Where:
• PP is the pressure,
• VV is the molar volume,
• TT is the temperature,
• aa, bb, and RR are constants.
We are tasked with finding the dimensions of the constant aa.
Step 1: Analyze the dimensions of each term in the equation.
1. Pressure (P): The dimension of pressure is given by:
[P]=MLT2[P] = \frac{M}{L T^2}
2. Molar volume (V): The molar volume has the dimension of volume per mole:
[V]=L3[V] = L^3
3. Temperature (T): The dimension of temperature is:
[T]=Θ(where Θ represents the dimension of temperature)[T] = \Theta \quad (\text{where} \, \Theta \, \text{represents the dimension of temperature})
4. Gas constant (R): The gas constant RR has the dimensions:
[R]=ML2T2Θ[R] = \frac{M L^2}{T^2 \Theta}
5. Constant aa: We need to find the dimensions of aa.
Step 2: Rearrange the equation to find the dimensions of aa.
The term inside the parentheses, P+aV2P + \frac{a}{V^2}, must have the dimension of pressure [P]=MLT2[P] = \frac{M}{L T^2}. So, the dimensions of aV2\frac{a}{V^2} should also be MLT2\frac{M}{L T^2}.
The term V2V^2 has dimensions of (L3)2=L6(L^3)^2 = L^6. Therefore, we can write:
[aV2]=[a]L6\left[ \frac{a}{V^2} \right] = \frac{[a]}{L^6}
Equating the dimensions of aV2\frac{a}{V^2} to the dimensions of pressure:
[a]L6=MLT2\frac{[a]}{L^6} = \frac{M}{L T^2}
Now solve for the dimensions of aa:
[a]=ML7T−2[a] = M L^7 T^{-2}
Step 3: Conclusion
The dimensions of aa are ML7T−2M L^7 T^{-2}.
This matches the choice (A): ML5T−2M L^5 T^{-2}.
Thus, the correct answer is (A).
