During random motion, gas molecules do not interact with each other . Hence Potential energy =0 ...... Pls explain why P.E = 0? P.E is stored form of energy,right?


Question and Answers Forum

All Questions Topic List
Heat and Theromdynamics Questions
Previous in All Question Next in All Question
Previous in Heat and Theromdynamics Next in Heat and Theromdynamics
Question Number 47314 by rahul 19 last updated on 08/Nov/18

$D u r i n g r a n d o m m o t i o n, g a s m o l e c u l e s$ $d o n o t i n t e r a c t w i t h e a c h o t h e r .$ $H e n c e P o t e n t i a l e n e r g y = 0 . . . . . .$ $P l s e x p l a i n w h y P . E = 0 ?$ $P . E i s s t o r e d f o r m o f e n e r g y, r i g h t ?$
Commented by rahul 19 last updated on 08/Nov/18

$t h i s i s o n e o f t h e p o s t u l a t e o f K . T . G . .$
	Answered by ajfour last updated on 08/Nov/18

	$d i s t a n c e b e t w e e n m o l e c u l e s i s$ $l a r g e, P i s l e s s a n d / o r t e m p e r a t u r e$ $h i g h, m o l e c u l e s n e u t r a l, p o t e n t i a l$ $e n e r g y t h e r e f o r e n e g l i g i b l e$ $c o m p a r e d t o k i n e t i c e n e r g y .$
	Commented by rahul 19 last updated on 08/Nov/18

	$S i r, {w h a t}^{'} s d i f f . b e t w e e n$ $g a s m o l e c u l e s i n t e r a c t w i t h e a c h o t h e r .$ $& g a s m o l e c u l e s c o l l i d e w i t h e a c h o t h e r .$
	Commented by ajfour last updated on 08/Nov/18

	$d u r i n g c o l l i s i o n, t h i s i s e v e n a t$ $s o m e d i s t a n c e (w h e n c h a r g e d b o d i e s$ $c o n c e r n e d), p o t e n t i a l e n e r g y r i s e s$ $c o n s i d e r a b l y b u t a f t e r c o l l i s i o n$ $r e t u r n s b a c k t o n e l i g i b l e v a l u e$ $f o r i d e a l g a s a t o m s . S o a s f a r a s$ $e l e c t r o s t a t i c i n t e r a c t i o n i s$ $c o n c e r n e d, t h i s t a k e s p l a c e o n l y$ $d u r i n g n o n - c o n t a c t c o l l i s i o n s;$ $c o n t a c t c o l l i s i o n s m a y r e s u l t i n$ $a c h e m i c a l r e a c t i o n .$
	Commented by rahul 19 last updated on 08/Nov/18
	thanks sir ��
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum