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Question: Boyle’s law may be expressed as \( {\left( {\dfrac{{{\text{dP}}}}{{{\text{dV}}}}} \right)_{\text{T}}...

Boyle’s law may be expressed as (dPdV)T =  - kV2{\left( {\dfrac{{{\text{dP}}}}{{{\text{dV}}}}} \right)_{\text{T}}}{\text{ = }}\dfrac{{{\text{ - k}}}}{{{{\text{V}}^{\text{2}}}}} . If true enter 1 or else 0.

Explanation

Solution

In the above question, we have to check whether Boyle's law can be written as the equation given. For this, we have to see what Boyle's law is and then we will differentiate the equation to check whether the obtained equation is equal to the equation given.
Formula Used: V1P{\text{V}} \propto \dfrac{{\text{1}}}{{\text{P}}}

Complete step by step solution:
We know that Boyle’s law describes how the pressure of a gas tends to increase as the volume of the container decreases. In other words, it states that volume of a gas is inversely proportional to the pressure applied to the gas provided the temperature of the surrounding remains constant.
Mathematically, it can be written as:
V1P{\text{V}} \propto \dfrac{{\text{1}}}{{\text{P}}}
Where V is the volume of the gas and P is pressure applied to the gas at a particular temperature.
Introducing a constant of proportionality, k in the equation, we get:
PV = k{\text{PV = k}}
Rearranging the above equation, we get:
P = kV{\text{P = }}\dfrac{{\text{k}}}{{\text{V}}}
Differentiating the above equation with respect to volume V at a constant temperature T, we get:
(dPdV)T =  - kV2{\left( {\dfrac{{{\text{dP}}}}{{{\text{dV}}}}} \right)_{\text{T}}}{\text{ = }}\dfrac{{{\text{ - k}}}}{{{{\text{V}}^{\text{2}}}}}
And the obtained equation is the same as the given equation and hence, it is true.

The value to be entered is 1.

Note:
Boyle's law is often used as part of an explanation on how the breathing system works in the human body. It is commonly used to explain how the lung volume can be increased or decreased and thereby causing a relatively lower or higher air pressure within them (which in context with Boyle's law).