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Question-52992




Question Number 52992 by prakash jain last updated on 16/Jan/19
Commented by prakash jain last updated on 16/Jan/19
Correct answer is ((4−R)/(2R)), NTA has given  ((4+R)/(2R)). This question is also from  9th Jan morning JEE main exam.
Correctansweris4R2R,NTAhasgiven4+R2R.Thisquestionisalsofrom9thJanmorningJEEmainexam.
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19
P_A =N_A (((RT)/V))  P_B =N_B (((RT)/V))  P=P_A +P_B =((RT)/V)(N_A +N_B )  N_A +N_B =((PV)/(RT))  N_B =((PV)/(RT))−N_A   N_B =((200×10)/(R×1000))−0.5  N_B =(2/R)−(1/2)  N_B =((4−R)/(2R))
PA=NA(RTV)PB=NB(RTV)P=PA+PB=RTV(NA+NB)NA+NB=PVRTNB=PVRTNANB=200×10R×10000.5NB=2R12NB=4R2R
Answered by afachri last updated on 16/Jan/19
Answered by FelipeLz last updated on 26/Oct/21
PV = nRT  2.10^2 ∙10 = ((1/2)+x)∙R∙10^3   2 = ((1/2)+x)∙R  2 = ((1/2)+x)∙R  (2/R) = (1/2)+x  x = (2/R)−(1/2)  x = ((4−R)/(2R))
PV=nRT2.10210=(12+x)R1032=(12+x)R2=(12+x)R2R=12+xx=2R12x=4R2R

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