Figure shows, the adiabatic curve on a log T and log V scale performed on ideal gas. The gas is

Question

Figure shows, the adiabatic curve on a log T and log V scale performed on ideal gas. The gas is (A) Monatomic (B) Diatomic (C) Polyatomic (D) Mixture of monatomic and diatomic

Tardigrade Admin · Accepted Answer

T V&gamma;-1=K log T+(&gamma;-1) log V=0 log T=-(&gamma;-1) log V y=-(&gamma;-1) x ⇒ -(&gamma;-1)=-(2/3) (y/x)=-(&gamma;-1)= slope =(2-4/4-1) &gamma;=(5/3) ∴ Monoatomic.

Q. Figure shows, the adiabatic curve on a log $T$ and $lo g V$ scale performed on ideal gas. The gas is

Monatomic

Diatomic

Polyatomic

Mixture of monatomic and diatomic

$T V^{γ - 1} = K$
$lo g T + (γ - 1) lo g V = 0$
$lo g T = - (γ - 1) lo g V$
$y = - (γ - 1) x$
$\Rightarrow - (γ - 1) = - \frac{2}{3}$
$\frac{y}{x} = - (γ - 1) =$ slope $= \frac{2 - 4}{4 - 1}$
$γ = \frac{5}{3}$
$∴$ Monoatomic.

Q. Figure shows, the adiabatic curve on a log T and logV scale performed on ideal gas. The gas is