ISITEKS: DIFERENSIAL EKSAK

Diketahui:

$\large \dpi{120} \bg_black xdy+ydx=d(xy)$

$\large \dpi{120} \bg_black ydy=d(\frac{1}{2}y^{2})$

$\large \dpi{120} \bg_black xdx=d(\frac{1}{2}x^{2})$

Apakah:

$\large \dpi{120} \bg_black 2xdx+3dy=diferensial\ eksak?$

$\large \dpi{120} \bg_black xdy+ydx=diferensial\ eksak?$

$\large \dpi{120} \bg_black xdx+ydy=diferensial\ eksak?$

Diferensial eksak $\large \dpi{120} \equiv$ diferensial dari suatu fungsi
Diferensial bukan eksak $\large \dpi{120} \equiv$ diferensial yang bukan diferensial eksak

Cara ampuh membedakan diferensial eksak dengan tidak eksak, dari $\large \dpi{120} \bg_black dz=M(x,y)dx+N(x,y)dy$

Jika $\large \dpi{120} \bg_black (\frac{\partial M(x,y)}{\partial y})_{x}=(\frac{\partial N(x,y)}{\partial x})_{y}$

maka dz adalah diferensial eksak
sedangkan jika $\large \dpi{120} \bg_black \frac{\partial M(x,y)}{\partial y}\neq \frac{\partial N(x,y)}{\partial x}$
maka dz adalah diferensial tidak eksak

Jawaban:

1. $\large \dpi{120} \bg_black 2dx+3dy=\ diferensial\ eksak?$
$\large \dpi{120} \bg_black \rightarrow d\left ( 2x+3y \right )=diferensial\ fungsi=diferensial\ eksak$

2. $\large \dpi{120} \bg_black xdy+ydx=d(xy)=diferensial\ fungsi=diferensial\ eksak$

3. $\large \dpi{120} \bg_black xdx+ydy=d(\frac{1}{2}x^{2})+d(\frac{1}{2}y^{2})$

   $\large \dpi{120} \bg_black =d\left ( \frac{1}{2} x^{2}+\frac{1}{2}y^{2}\right )=diferensial\ fungsi=diferensial\ eksak$

Soal:
Gunakan relasi silang untuk memeriksa keeksakan diferensial-diferensial berikut:

$\large \dpi{120} \bg_black a)2dx+3dy$

$\large \dpi{120} \bg_black b)ydx+xdy$

$\large \dpi{120} \bg_black c)xdx+ydy$

$\large \dpi{120} \bg_black d)xdy-ydx$

Jawab:

a)   $\large \dpi{120} \bg_black 2dx+3dy\rightarrow M=2 \ \ N=3$

$\large \dpi{120} \bg_black \frac{\partial M}{\partial y}=\frac{\partial 2}{\partial y}=0\ \ \ \ \ \ \ \frac{\partial N}{\partial x}=\frac{\partial 3}{\partial x}=0$

$\large \dpi{120} \bg_black \rightarrow \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\rightarrow 2dx+3dy \ \ adalah\ \ diferensial\ eksak$

b) $\large \dpi{120} \bg_black ydx+xdy\rightarrow M=y\ \ \ \ \ N=x$

$\large \dpi{120} \bg_black \frac{\partial M}{\partial y}=\frac{\partial y}{\partial y}=1\ \ \ \ \ \ \ \frac{\partial N}{\partial x}=\frac{\partial x}{\partial x}=1$

karena $\large \dpi{120} \bg_black \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ maka $\large \dpi{120} \bg_black xdy+ydx=diferensial\ eksak$

c) $\large \dpi{120} \bg_black xdx+ydy\rightarrow M=x\ \ \ \ \ \ \ \ N=y$

$\large \dpi{120} \bg_black \frac{\partial M}{\partial y}=\frac{\partial x}{\partial y}=\frac{\partial tetap}{\partial y}=0$

$\large \dpi{120} \bg_black \frac{\partial N}{\partial x}=\frac{\partial y}{\partial x}=\frac{\partial tetap}{\partial x}=0$

   $\large \dpi{120} \bg_black \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

$\large \dpi{120} \bg_black \rightarrow xdx+ydy\ adalah\ diferensial\ eksak$

d) $\large \dpi{120} \bg_black xdy-ydx\rightarrow M=-y\ \ \ \ \ \ \ \ \ N=x$

$\large \dpi{120} \bg_black \frac{\partial M}{\partial y}=-\frac{\partial y}{\partial y}=-1$
   $\large \dpi{120} \bg_black \frac{\partial N}{\partial x}=\frac{\partial x}{\partial x}=1$

$\large \dpi{120} \bg_black \frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}$

$\large \dpi{120} \bg_black \rightarrow xdy-ydx\ adalah\ bukan\ diferensial\ eksak$

MATERI GAS IDEAL

Diketahui:

Gas ideal dengan $\large \dpi{120} \bg_black pV=nRT$ memiliki $\large \dpi{120} \bg_black \left ( \frac{\partial U}{\partial V} \right )_{T}=0$

$\large \dpi{120} \bg_black dU=\left ( \frac{\partial U}{\partial T} \right )_{V}dT+\left ( \frac{\partial U}{\partial V}_\right )_{T}dV$ ,
U adalah fungsi atau dU=diferensial eksak
$\large \dpi{120} \bg_black \left ( \frac{\partial }{\partial V}\left ( \frac{\partial U}{\partial T} \right )_{V} \right )_{T}=\left [ \frac{\partial }{\partial T}\left ( \frac{\partial U}{\partial V} \right )_{T} \right ]$

Buktikan:

$\large \dpi{120} \bg_black dS=\frac{dU+pdV}{T}$ pada n tetap adalah diferensial eksak

Bukti :

Gas ideal    $\large \dpi{120} \bg_black dU=\left ( \frac{\partial U}{\partial T} \right )_{V}dT+\left ( \frac{\partial U}{\partial V} \right )_{T}dV$

   $\large \dpi{120} \bg_black =\left ( \frac{\partial U}{\partial T} \right )_{V}dT+0\ dV$

   $\large \dpi{120} \bg_black dU=\left ( \frac{\partial U}{\partial T} \right )_{V}dT$

$\large \dpi{120} \bg_black dS=\frac{\left ( \frac{\partial U}{\partial T} \right )_{V}dT+\frac{nRT}{V}dV}{T}=\frac{1}{T}\left ( \frac{\partial U}{\partial T} \right )_{V}dT+\frac{nR}{V}dV$

$\large \dpi{120} \bg_black M=\frac{1}{T}\left ( \frac{\partial U}{\partial T} \right )_{V}$

$\large \dpi{120} \bg_black N=\frac{nR}{V}\rightarrow\left ( \frac{\partial N}{\partial T} \right )_{V}=\left ( \frac{\partial }{\partial T}\ \frac{nR}{V} \right )_{V}=\frac{nR}{V}\left ( \frac{\partial \cdot 1}{\partial T} \right )=0$

$\large \dpi{120} \bg_black \left ( \frac{\partial M}{\partial V} \right )_{T}=\left ( \frac{\partial }{\partial V}\ \frac{1}{T} \left ( \frac{\partial U}{\partial T} \right )_{V}\right )_{T}$

   $\large \dpi{120} \bg_black =\frac{1}{T}\left [ \frac{\partial }{\partial V}\left ( \frac{\partial U}{\partial T} \right )_{V} \right ]_{T}$

   $\large \dpi{120} \bg_black =\frac{1}{T}\left ( \frac{\partial }{\partial T}\left ( \frac{\partial U}{\partial V} \right )_{T} \right )_{V}=\frac{1}{T}\left ( \frac{\partial }{\partial T}\cdot 0 \right )_{V}=0$

$\dpi{120} \bg_black \left ( \frac{\partial M}{\partial V} \right )_{T}=\left ( \frac{\partial N}{\partial T} \right )_{V}\ \ \ maka\ dS=diferensial\ eksak$

ISITEKS

Senin, 04 Maret 2013

DIFERENSIAL EKSAK

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