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Question: The pressure P and the volume v of a gas are connected by the relation \(p{{v}^{1.4}}=const\). Find ...

The pressure P and the volume v of a gas are connected by the relation pv1.4=constp{{v}^{1.4}}=const. Find the percentage error in p corresponding to a decrease of 12%\dfrac{1}{2}\% in v.

Explanation

Solution

Here, we can first represent the percentage error in v in terms of p and then we will try to find the percentage error in p. We will take the help of differentiation to find the error percentage as relation is in exponent form and we will use the formula:
Percentage error= !!∣!! accepted value-experimental value !!∣!! accepted value×10%\dfrac{\text{ }\\!\\!|\\!\\!\text{ accepted value-experimental value }\\!\\!|\\!\\!\text{ }}{\text{accepted value}}\times 10\%

Complete step-by-step answer :
The result of every measurement by any measuring instrument contains some uncertainty. This uncertainty is called the error. It is the difference between true value and measured value of the quantity and is also known as error of measurement. Basically, there are two types of errors in physics, random errors and systematic errors.
Percent error or percentage error expresses as a percentage the difference between an approximate or measured value and an exact or known value. It is used in science to report the difference between a measured or experimental value and a true value.
Percent error is the difference between a measured and known value divided by the known value and then multiplied by 100%.
Percentage error =  !!∣!! accepted value-experimental value !!∣!! accepted value×10%\dfrac{\text{ }\\!\\!|\\!\\!\text{ accepted value-experimental value }\\!\\!|\\!\\!\text{ }}{\text{accepted value}}\times 10\%
Since, the relation given to us is:
pv1.4=constp{{v}^{1.4}}=const
Here, we will use the product rule of differentiation to differentiate the above equation. The product rule of differentiation for two functions u(x) and v(x) is given as:
d(u×v)dx=u.dvdx+v.dudx\dfrac{d\left( u\times v \right)}{dx}=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}.
Also, the differentiation of xn{{x}^{n}} is given as d(xn)dx=n×x(n−1)\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\times {{x}^{\left( n-1 \right)}}.
So, on applying these two rules, we get:
p×1.4v1.4−1×Δv+v1.4×Δp=0 ⇒p×1.4×v0.4.Δv=−Δp.v1.4 ⇒1.4×v0.4v1.4×Δv=−Δpp ⇒1.4×1v1.4−0.4×Δv=−Δpp ⇒1.4×Δvv=−Δpp \begin{aligned} & p\times 1.4{{v}^{1.4-1}}\times \Delta v+{{v}^{1.4}}\times \Delta p=0 \\\ & \Rightarrow p\times 1.4\times {{v}^{0.4}}.\Delta v=-\Delta p.{{v}^{1.4}} \\\ & \Rightarrow \dfrac{1.4\times {{v}^{0.4}}}{{{v}^{1.4}}}\times \Delta v=\dfrac{-\Delta p}{p} \\\ & \Rightarrow 1.4\times \dfrac{1}{{{v}^{1.4-0.4}}}\times \Delta v=\dfrac{-\Delta p}{p} \\\ & \Rightarrow 1.4\times \dfrac{\Delta v}{v}=\dfrac{-\Delta p}{p} \\\ \end{aligned}
On multiplying both sides by 100 to get percentage error, we get:
1.4×Δvv×100=−Δpp×100 ⇒1.4×12=−Δpp×100 \begin{aligned} & 1.4\times \dfrac{\Delta v}{v}\times 100=\dfrac{-\Delta p}{p}\times 100 \\\ & \Rightarrow 1.4\times \dfrac{1}{2}=\dfrac{-\Delta p}{p}\times 100 \\\ \end{aligned}
Therefore, ∣Δpp×100∣=0.7%\left| \dfrac{\Delta p}{p}\times 100 \right|=0.7\%
Hence, the percentage error in p corresponding to a decrease of 12%\dfrac{1}{2}\% in v is = 0.7%.

Note :
Students should note here that it is not always necessary to take the absolute value but in general cases, we take modulus to find the absolute value of error. In case the percentage error comes to be negative, it means that the obtained value or the new value is less than the actual value.The calculations must be done carefully to avoid unnecessary mistakes.