Question

The pressure p and the volume v of a gas are connected by the relation pv^1.4 = const. Find the percentage error in p corresponding to a decrease of 1/2% in v .

Answer 1

\[\text { We have }\]

\[p v^{1 . 4} = \text { constant} = k \left( \text { say } \right)\]

\[\text { Taking log on both the sides, we get }\]

\[\log \left( p v^{1 . 4} \right) = \log k\]

\[ \Rightarrow \log p + 1 . 4 \log v = \log k\]

\[\text { Differentiating both the sides w . r . t . x, we get }\]

\[\frac{1}{p}\frac{dp}{dv} + \frac{1 . 4}{v} = 0\]

\[ \Rightarrow \frac{dp}{p} = \frac{- 1 . 4 dv}{v}\]

\[\text { Now, dp } = \frac{dp}{dv}dv = \frac{- 1 . 4p}{v}dv\]

\[ \Rightarrow \frac{dp}{p} \times 100 = - 1 . 4\left( \frac{dv}{v} \times 100 \right) = - 1 . 4 \times \left( \frac{- 1}{2} \right) = 0 . 7 \left[ \text { Since we are given} \frac{1}{2} \% \text { decrease in} v \right]\]

\[\text { Hence, the error in p is } 0 . 7 \% .\]