Question 328: Thermodynamics & KTG

Question

Which one is the correct option for the two different thermodynamic processes?

Accepted Answer

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.

Answer

(a) only

Answer

(c) and (d)

Answer

(b) and (c)

Answer

(c) and (a)

Thermodynamics & KTG

Which one is the correct option for the two different thermodynamic processes?

Choose the correct answer: