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  • What is the value of <img alt="\mathrm{-\frac{1}{V}\left(\frac{d V}{d p}\right)_{T, n}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B-%5Cfrac%7B1%7D%7BV%7D%5Cleft%28%5Cfrac%7Bd

What is the value of \mathrm{-\frac{1}{V}\left(\frac{d V}{d p}\right)_{T, n}} at 0.2 ban?
 

Option: 1

6 bap ^{-1}


Option: 2

a b a r b^{-1}


Option: 3

5 \mathrm{bar}^{-1}


Option: 4

4 \mathrm{bar}^{-1}


Answers (1)

best_answer

This is a case of Boyle's law under isothermal conditions.

\begin{aligned} &\mathrm{ V \alpha \frac{1}{p} \Rightarrow V=\frac{k}{p} }\\ &\mathrm{ \therefore\left(\frac{d V}{d p}\right)=-\frac{k}{p^2}=\frac{-p v}{p^2}=\frac{-V}{p} }\\ &\mathrm{ \therefore-\frac{1}{V}\left(\frac{d V}{d p}\right)=-\frac{1}{p}=\frac{1}{0.2}=5 \mathrm{bar}^{-1}}\\ \end{aligned}
option (c) correct.

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Rishabh

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