Question

A piston/cylinder contains water at $500^\circ C$ , $3\,MPa$. It is cooled in a polytropic process to $200^\circ C$, $1\,MPa$ .Find the polytropic exponent $(\eta )$.
A. $1.81$
B. ${\text{9.19}}$
C. ${\text{1.9}}$
D. ${\text{9.1}}$

Answer 1

In order to solve this question, we should know about the polytropic process along with their formula. Here we are given two situations: we have to find both the cases then we have to equate them to solve for the polytropic exponent $(\eta )$ . We have to convert all the units in conventional form for example: Celsius to kelvin

Formula required:
$P{V^\eta } = C$
We can also convert this formula to:
$\dfrac{{{T^\eta }}}{{{P^{\eta - 1}}}} = C$
Here, $P$ stands for pressure, $V$ stands for volume, $(\eta )$ stands for polytropic exponent, $C$ stands for constant and $T$ stands for temperature in kelvin.
$k = 273 + C^\circ$
We will use this formula to change from Celsius to kelvin.

Complete step by step answer:
Polytropic process is a thermodynamic process which follows the relation $P{V^\eta } = C$
Here $(\eta )$ stands for polytropic index or exponent ,when $(\eta )$ changes the type of process is changed. When $(\eta ) = 0$ process is an isobaric process , for $(\eta ) = \infty$ process is isochoric process and for $(\eta ) = 1$ process is isothermal process. $P{V^\eta } = C$ we cannot use this formula in this question because we are not given the volume so instead of this we will use this formula $\dfrac{{{T^\eta }}}{{{P^{\eta - 1}}}} = C$ as in question we given temperature.

Case 1:
${T_1}$ = $500^\circ C$
$\Rightarrow {P_1}$ = $3MPa$
${T_1}$ is the temperature in case 1 it is given in Celsius so we will change in kelvin
$k = 273 + C^\circ$
$\Rightarrow k =273 + 500^\circ$
$\Rightarrow k =773$
Case 2:
${T_2}$ = $200^\circ C$
$\Rightarrow {P_2}$ = $1MPa$
${T_2}$ is the temperature in case 2 it is given in Celsius so we will change in kelvin
$k = 273 + C^\circ$
$\Rightarrow k =273 + 200^\circ$
$\Rightarrow k =473$

Now using the formula we will find the $\eta$
$\dfrac{{{T^\eta }}}{{{P^{\eta - 1}}}} = C$
Case 1: here the temperature is ${T_1}$ and pressure is ${P_1}$
$\dfrac{{T{1^\eta }}}{{P{1^{\eta - 1}}}} = C$
$\Rightarrow \dfrac{{{{773}^\eta }}}{{{3^{\eta - 1}}}} = C$
Case 2: here the temperature is ${T_2}$ and pressure is ${P_2}$
$\dfrac{{T{2^\eta }}}{{P{2^{\eta - 1}}}} = C$
$\Rightarrow \dfrac{{{{473}^\eta }}}{{{1^{\eta - 1}}}} = C$
Now we will equate both the cases as both are equal to constant
$\dfrac{{{{773}^\eta }}}{{{3^{\eta - 1}}}} = \dfrac{{{{473}^\eta }}}{{{1^{\eta - 1}}}}$

Now we will take all pressure at right side and volume at left
$\dfrac{{{{773}^\eta }}}{{{{473}^\eta }}} = \dfrac{{{3^{\eta - 1}}}}{{{1^{\eta - 1}}}}$
$\Rightarrow \dfrac{{{{773}^\eta }}}{{{{473}^\eta }}} = \dfrac{{{3^{\eta - 1}}}}{{{1^{\eta - 1}}}}$
$\Rightarrow {\left( {\dfrac{{773}}{{473}}} \right)^\eta } = {3^{\eta - 1}}$
$\Rightarrow \dfrac{{{3^{\eta - 1}}}}{{{1^{\eta - 1}}}}$ becomes ${3^{\eta - 1}}$ because ${1^{\eta - 1}}$ becomes 1
Now we will take log both the sides
$\eta {\text{ ln}}\left( {\dfrac{{773}}{{473}}} \right) = (\eta - 1){\text{ ln(}}3)$
Values of logs
${\text{ ln}}\left( {\dfrac{{773}}{{473}}} \right) = 0.4911$
$\Rightarrow {\text{ln(}}3) = 1.098$

Now putting the values of log
$\eta {\text{ }}\left( {0.4911} \right) = (\eta - 1){\text{ (1}}{\text{.098)}}$
$\Rightarrow \dfrac{\eta }{{(\eta - 1)}}{\text{ }} = {\text{ }}\dfrac{{{\text{(1}}{\text{.098)}}}}{{\left( {0.4911} \right)}}$
Now we will reciprocal so as to simplify
$\dfrac{{\eta - 1}}{\eta }{\text{ }} = {\text{ }}\dfrac{{{\text{(0}}{\text{.4911)}}}}{{\left( {{\text{1}}{\text{.098}}} \right)}}$
$\Rightarrow 1 - \dfrac{1}{\eta }{\text{ }} = 0.44$
$\Rightarrow \dfrac{1}{\eta }{\text{ }} = 0.56$
$\therefore \eta = 1.81$

Hence option (A) is correct.

Note: Most of the people make mistakes in this type of question by finding the volume from pressure and temperature but instead of that we should use the derived formula in which we can directly put pressure and temperature.Along with this people forget to convert celsius into kelvin which further leads to incorrect answers.