Question Number 52992 by prakash jain last updated on 16/Jan/19
Commented by prakash jain last updated on 16/Jan/19
$$\mathrm{Correct}\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{4}−\mathrm{R}}{\mathrm{2R}},\:\mathrm{NTA}\:\mathrm{has}\:\mathrm{given} \\ $$$$\frac{\mathrm{4}+\mathrm{R}}{\mathrm{2R}}.\:\mathrm{This}\:\mathrm{question}\:\mathrm{is}\:\mathrm{also}\:\mathrm{from} \\ $$$$\mathrm{9th}\:\mathrm{Jan}\:\mathrm{morning}\:\mathrm{JE}{E}\:\mathrm{main}\:\mathrm{exam}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19
$${P}_{{A}} ={N}_{{A}} \left(\frac{{RT}}{{V}}\right) \\ $$$${P}_{{B}} ={N}_{{B}} \left(\frac{{RT}}{{V}}\right) \\ $$$${P}={P}_{{A}} +{P}_{{B}} =\frac{{RT}}{{V}}\left({N}_{{A}} +{N}_{{B}} \right) \\ $$$${N}_{{A}} +{N}_{{B}} =\frac{{PV}}{{RT}} \\ $$$${N}_{{B}} =\frac{{PV}}{{RT}}−{N}_{{A}} \\ $$$${N}_{{B}} =\frac{\mathrm{200}×\mathrm{10}}{{R}×\mathrm{1000}}−\mathrm{0}.\mathrm{5} \\ $$$${N}_{{B}} =\frac{\mathrm{2}}{{R}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${N}_{{B}} =\frac{\mathrm{4}−{R}}{\mathrm{2}{R}} \\ $$
Answered by afachri last updated on 16/Jan/19
Answered by FelipeLz last updated on 26/Oct/21
$${PV}\:=\:{nRT} \\ $$$$\mathrm{2}.\mathrm{10}^{\mathrm{2}} \centerdot\mathrm{10}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}+{x}\right)\centerdot{R}\centerdot\mathrm{10}^{\mathrm{3}} \\ $$$$\mathrm{2}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}+{x}\right)\centerdot{R} \\ $$$$\mathrm{2}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}+{x}\right)\centerdot{R} \\ $$$$\frac{\mathrm{2}}{{R}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}+{x} \\ $$$${x}\:=\:\frac{\mathrm{2}}{{R}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}\:=\:\frac{\mathrm{4}−{R}}{\mathrm{2}{R}} \\ $$