Question

For an isothermal expansion of a perfect gas, the value of $\dfrac{{\Delta P}}{P}$ is equal to
A. ${\gamma ^{\dfrac{1}{2}}} \cdot \dfrac{{\Delta V}}{V}$
B. $ - \dfrac{{\Delta V}}{V}$
C. $ - \gamma \cdot \dfrac{{\Delta V}}{V}$
D. ${\gamma ^2} \cdot \dfrac{{\Delta V}}{V}$

Answer 1

Hint:
In an Isothermal process in a thermodynamic system, the temperature is constant and at a constant temperature, the value of $\dfrac{{\Delta P}}{P}$ for a perfect gas can be determined by using an ideal gas equation $PV = nRT$ and differentiating and simplifying it and then, identify the correct option for the given problem.

Formula used:
An Ideal-Gas Equation, $PV = nRT$
and product rule of differentiation $\dfrac{d}{{dx}}\left( {I \cdot II} \right) = I \cdot \dfrac{d}{{dx}}\left( {II} \right) + II \cdot \dfrac{d}{{dx}}\left( I \right)$




Complete step by step solution:

An isothermal Process in thermodynamics is defined as the process during which the temperature $T$ of a system remains constant that’s why it is also referred to as a constant-temperature process.
In an Isothermal process, $\,Change{\text{ }}in{\text{ }}Temperature = \Delta T = 0$
i.e., $T = constant$
Now, we know that an ideal-gas equation can be stated as: -
$PV = nRT$
In an Isothermal expansion, $T = constant$
$ \Rightarrow PV = nR(constant)$
As $n\,and\,R$ are constant for given ideal gas.
$ \Rightarrow PV = constant = k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)$
By differentiating and using the product rule $\dfrac{d}{{dx}}\left( {I \cdot II} \right) = I \cdot \dfrac{d}{{dx}}\left( {II} \right) + II \cdot \dfrac{d}{{dx}}\left( I \right)$ in equation $(1)$, we get
$ \Rightarrow PdV + VdP = 0$ $\left\{ {\therefore \dfrac{d}{{dx}}(constant) = 0} \right\}$
$ \Rightarrow \dfrac{{dP}}{P} = - \dfrac{{dV}}{V}$ (This is for small change)
For a large change, we can also rewrite the above expression as: -
$ \Rightarrow \dfrac{{\Delta P}}{P} = - \dfrac{{\Delta V}}{V}$
Thus, for an isothermal expansion of a perfect gas, the value of $\dfrac{{\Delta P}}{P}$ is equal to $ - \dfrac{{\Delta V}}{V}$.
Hence, the correct option is (B) $ - \dfrac{{\Delta V}}{V}$ .





Note:
In this problem, to determine the value of $\dfrac{{\Delta P}}{P}$ for an isothermal expansion of a perfect gas, we have to apply the condition of isothermal change, i.e., $\Delta T = 0$ in ideal gas equation $PV = nRT$ , and then differentiate it and then analyze each given option carefully to give an accurate answer with exact reasons.