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

$$\text{ We have }$$

$$p{v}^{1.4}=\text{ constant}=k\left(\text{ say }\right)$$

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

$$\log \left(p{v}^{1.4}\right)=\log k$$

$$\Rightarrow \log p+1.4\log v=\log k$$

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

$$\frac{1}{p}\frac{dp}{dv}+\frac{1.4}{v}=0$$

$$\Rightarrow \frac{dp}{p}=\frac{-1.4dv}{v}$$

$$\text{ Now, dp }=\frac{dp}{dv}dv=\frac{-1.4p}{v}dv$$

$$\Rightarrow \frac{dp}{p}\times 100=-1.4(\frac{dv}{v}\times 100)=-1.4\times \left(\frac{-1}{2}\right)=0.7\left[\text{ Since we are given}\frac{1}{2}\mathrm{\%}\text{ decrease in}v\right]$$

$$\text{ Hence, the error in p is }0.7\mathrm{\%}.$$