JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2024
- Hard
- MCQ
The equation of state of a real gas is given by \(\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}\), where \(\mathrm{P}, \mathrm{V}\) and \(\mathrm{T}\) are pressure. volume and temperature respectively and \(R\) is the universal gas constant. The dimensions of \(\frac{a}{b^2}\) is similar to that of :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\mathrm{P}\)
Step-by-step Solution
Detailed explanation
\( {[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right] \Rightarrow[\mathrm{a}]=\left[\mathrm{PV}^2\right]} \)
\( \text { And }[\mathrm{V}]=[\mathrm{b}] \)
\( \frac{[\mathrm{a}]}{\left[\mathrm{b}^2\right]}=\frac{\left[\mathrm{PV}^2\right]}{\left[\mathrm{V}^2\right]}=[\mathrm{P}]\)
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