Question

Boyle’s law may be expressed as ${\left( {\dfrac{{dP}}{{dV}}} \right)_T} =
\dfrac{K}{V}$
If true enter 1 else 0

Answer 1

Hint: We know that Boyle’s law states, volume of a given mass of a gas is inversely proportional to its pressure, keeping the temperature constant. From this we can define the relation we required.

Complete step by step solution:
By the definition of Boyle’s law, we can say that volume is inversely proportional to pressure when temperature is kept constant.
So, mathematically
$V\; \propto \dfrac{1}{P}$ at constant temperature
We can also say that $PV = constant$
When we take ${P_{1,}}{V_1}$ as the initial pressure and volume respectively also ${P_{2,}}{V_2}$ as the final pressure and volume respectively we can write this as
${{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}$
 Earlier we said that $PV = constant$, let rewrite the equation as
$P = {\text{K}}{{\text{V}}^{ - 1}}$
Now, since the temperature is constant, we can differentiate pressure with respect to volume.
$\dfrac{{d{\text{P}}}}{{d{\text{V}}}} = \dfrac{{d\left( {K{V^{ - 1}}} \right)}}{{dV}}$
Since K is the constant we can take it outside
$\dfrac{{d{\text{P}}}}{{d{\text{V}}}} = {\text{K}}\dfrac{{d{V^{ - 1}}}}{{dV}}$
Since $\dfrac{{d\left( {\dfrac{1}{{\text{x}}}} \right)}}{{d{\text{x}}}} = - \dfrac{1}{{{x^2}}}$
Applying this we get
$\dfrac{{dP}}{{dV}} = - \dfrac{K}{{{V^2}}}$
Since we took the temperature as constant
${\left( {\dfrac{{dP}}{{dV}}} \right)_T} = - \dfrac{K}{{{V^2}}}$

Hence the given formula is false. Therefore, enter 0.

Note: The significance of Boyle’s law: It can be concluded that gases are compressible. Since the gas density is directly proportional to pressure, so more the gas is compressed, denser it becomes. The graph plotted for pressure against volume we can see a curve which indicates it is inversely proportional to it.
Now, let us look into an example of Boyle’s law, when a balloon is squeezed we can see the volume of the balloon decreases, this is accompanied by an increase in pressure. We can see that the more we press the balloon the more difficult it becomes due to the pressure exerted by the balloon. This is one of the examples of Boyle’s law.