JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 15)
Which one is the correct option for the two different thermodynamic processes?
_17th_March_Evening_Shift_en_15_1.png)
_17th_March_Evening_Shift_en_15_2.png)
_17th_March_Evening_Shift_en_15_3.png)
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(a) only
(c) and (d)
(b) and (c)
(c) and (a)
Explanation
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
P1 $$-$$ $$\gamma$$ 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.
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
P1 $$-$$ $$\gamma$$ 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.
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