Question 328: Thermodynamics & KTG

Question

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