Question

The minimum attainable pressure of ideal gas in the process $T = {T_0} + \alpha v2$ (where ${T_0}\,and\,\alpha $ are positive constant and v is the volume of one mole of gas) is
A. $5R\sqrt {\alpha {T_0}} $
B. $2R\sqrt {\alpha {T_0}} $
C. $2{T_0}\sqrt {\alpha R} $
D. $5{T_0}\sqrt {\alpha R} $

Answer 1

Hint: An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state and is amenable to analysis under statistical mechanics.

Complete step by step answer:
Given, $T = {T_0} + \alpha v2$….(1)
For $1\,mol$ of gas,
$PV = nRT$
$ \Rightarrow V = \dfrac{{RT}}{P}$
Substitute this value in equation 1
\[T = {T_0} + \alpha {\left( {\dfrac{{RT}}{P}} \right)^2}\]
\[ \Rightarrow T = {T_0} + \alpha \dfrac{{{R^2}{T^2}}}{{{P^2}}}\]
\[ \Rightarrow T{P^2} = {T_0}{P^2} + \alpha {R^2}{T^2}\]
\[ \Rightarrow P = \sqrt \alpha RT{(T - {T_0})^{ - \dfrac{1}{2}}}\]…(2)

Differentiate w.r.t ‘T’
$\dfrac{{dP}}{{dT}} = \sqrt \alpha T\left[ {{{(T - {T_0})}^{ - \dfrac{1}{2}}} - \dfrac{1}{2}T{{(T - {T_0})}^{ - \dfrac{3}{2}}}} \right]$
For minimum pressure $\dfrac{{dP}}{{dT}} = 0$
$ \Rightarrow 0 = \sqrt \alpha T\left[ {{{(T - {T_0})}^{ - \dfrac{1}{2}}} - \dfrac{1}{2}T{{(T - {T_0})}^{ - \dfrac{3}{2}}}} \right]$
Simplify
$T = 2{T_0}$
From equation 2
\[{P_{\min }} = \sqrt \alpha R2{T_0}{(2{T_0} - {T_0})^{ - \dfrac{3}{2}}}\]
\[ \therefore {P_{\min }} = 2R\sqrt {\alpha {T_0}} \]

Hence, the correct answer is option B.

Note: The ideal gas volume has negligible volume. The gas particles are equally sized and do not have intermolecular force (attraction or repulsion) with other gas particles. The gas particles move randomly in agreement with Newton's law of motion. The collisions between particles are elastic and their motion is frictionless.