Class 12th

Applying : ${\left(n_1+n_2\right)}_{initial}={\left(n_1+n_2\right)}_{final}$  

Assuming the system attains a final temperature of T (Such that 300 < T < 60)

(Heat lost by N₂ of container l) = (Heat gained by N₂ of container II)

$n_{1}Cm\left(300-T\right)=n_{2}Cm\left(T-60\right)$  

$\left(\frac{2.8}{28}\right)\left(300-T\right)=\frac{0.2}{28}\left(T-60\right)$  

14 (300 – T) = T – 60

$P=\left(\frac{3}{28}\right)*8.31*\frac{284}{3*10^{-3}}*10^{-5}bar=0.84287bar$

$P=84.28*10^{-2}bar$

$\approx 84*10^{-2}bar$  

Energy of emitted photon in 0.1 sec = 10^-4 J

$n*\frac{hc}{\lambda }=10^{-4}$

$\frac{n*6.63*10^{-34}*3*10^8}{1000*10^{-9}}=10^{-4}$

$n=5.02*10^{14}=50.2*10^{13}\approx 50*10^{13}$

$K_{700}=6.36*10^{-3}{s}^{-1}$

$K_{600}=x*10^{-6}{s}^{-1}$

E_a = 209 KJ/mol

$log\left(\frac{K_2}{K_1}\right)=\frac{-Ea}{2.303R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

$log{K}_{600}=-4.79$           

$K_{600}=10^{-4.79}=1.62*10^{-5}$

= 16.2 * 10^-6

x = 16 (the nearest integer)

Two optical isomers of ${\left[Cr{\left(C_2{O}_4\right)}_3\right]}^{3-}$ are shown :

$\eta =\frac{1}{4}or25\mathrm{\%}$

$\Delta T_{sink}=-58°C\left(iedecreased\right)$

$\eta \text{'}=\frac{1}{2}or50\mathrm{\%}$

T_{sin k} =?

As we know, the efficiency for a reversible engine is given by :

$\eta =1-\frac{T_2}{T_1}\left[T_2\to sinktemperature\right]$

On comparing both, we get

$T_2=\frac{3}{4}{T}_1\&\left(T_2-58°C\right)=\frac{T_1}{2}$

$\Rightarrow T_2=174°C$

$C{H}_{3}C{H}_{2}C{H}_{2}Br\stackrel{Mg}{\to }C{H}_{3}C{H}_{2}C{H}_{2}MgBr\stackrel{C{O}_2}{\to }$

$C{H}_{3}C{H}_{2}C{H}_{2}COOMgBr\stackrel{H_3{O}^{+}}{\to }C{H}_{3}C{H}_{2}C{H}_{2}COOH\left(butanoicacid\right)$

dist l1 l2 $=2\text{exception 25:}a$

= 2 * 2.3 a

= 4.6 a

O₂F₂ oxidises plutonium to PuF₆ and the reaction is used in removing plutonium as PuF₆ from spent nuclear fuel.