JEE Main 12 Jan 2019 Morning Question 4
Question: Two solids dissociate as follows
$ {A _{(s)}}{B _{(g)}}+{C _{(g)}};K _{P1}=x,atm^{2} $ $ {D _{(s)}}{C _{(g)}}+{E _{(g)}};K _{P2}=y,atm^{2} $ The total pressure when both the solids dissociate simultaneously is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ \sqrt{x+y}atm $
B) $ x^{2}+y^{2}atm $
C) $ 2( \sqrt{x+y} )atm $
D) $ (x+y)atm $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {A _{(s)}}\underset{P _1}{\mathop{{B _{(g)}}}},+\underset{P _1+P _2}{\mathop{{C _{(g)}}}}, $ $ {D _{(s)}}\underset{P _1+P _2}{\mathop{{C _{(g)}}}},+\underset{P _2}{\mathop{{E _{(g)}}}}, $
$ {K _{P _1}}=x=P _1(P _1+P _2)atm^{2},{K _{P _2}}=y=P _2(P _1+P _2)atm^{2} $
$ x+y={{(P _1+P _2)}^{2}}\Rightarrow P _1+P _2=\sqrt{x+y} $
$ P _{Total}=P _{B}+P _{C}+P _{E}=2(P _1+P _2)=2\sqrt{x+y}atm $
