JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2025
- Medium
- MCQ
The equation for real gas is given by \(\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}\), where \(\mathrm{P}, \mathrm{V}, \mathrm{T}\) and R are the pressure, volume, temperature and gas constant, respectively. The dimension of \(\mathrm{ab}^{-2}\) is equivalent to that of :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) Energy density
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & {\left[\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right](\mathrm{V}-\mathrm{b})=\mathrm{RT}} \\ & \therefore[\mathrm{a}]=[\mathrm{P}]\left[\mathrm{V}^2\right]=\mathrm{ML}^{-1} \mathrm{~T}^{-2} \mathrm{~L}^6=\mathrm{ML}^5 \mathrm{~T}^{-2} \\ & {[\mathrm{~b}]=[\mathrm{V}]=\mathrm{L}^3} \\ & {\left[\mathrm{ab}^{-2}\right]=\mathrm{ML}^5 \mathrm{~T}^{-2} \mathrm{~L}^{-6}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}}\end{aligned}\)
Dimension of energy density.
Dimension of energy density.
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