QJEEM21Phcb61: JEE Main 2021 Physics - Class 11 JEEM21Phcb61 Solution: Which one is the correct option for the two different…

Isothermal process means constant temperature which is only possible in graph (c) & (d) for adiabatic process

pv ^$\gamma$ = constant ........... (1)

$\because$ PV = nRT

$p$ $\propto$ ${T \over v}$

So, ${T \over v}$ v ^$\gamma$ = constant

Tv ^{$\gamma$ $-$ 1} = constant ......... (2)

Similarly,

$v \propto {T \over p}$

$P{\left( {{T \over P}} \right)^\gamma }$ = constant

P^{1 $-$ $\gamma$} T ^$\gamma$ = constant ......... (3)

$\because$ differentiating equation (3) w.r.t. temp.

${(P)^{1 - \gamma }}\gamma {(T)^{\gamma - 1}}dT + {(T)^\gamma }(1 - \gamma ){(P)^{1 - \gamma - 1}}dP = 0$

${{dP} \over {dT}} = {{(1 - \gamma ){T^\gamma }{P^{ - \gamma }}} \over {\gamma {{(P)}^{1 - \gamma }}{{(T)}^{\gamma - 1}}}} = {{(\gamma - 1)T} \over {\gamma P}}$

It gives (+ve) slope.