When the above given sample of $N_2{O}_4$ is heated to 327°C, 50% of “Hg” by weight is spilled out. $(P_{atm}=76cmHg)$ . Hence we can conclude that __________

Due to increase in temperature. and dissociation of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$. Hence $\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}$

 $\text{V\hspace{0.17em}}\alpha 1\therefore$(Area is uniform)

Initial $P_1=76+76=152mhg$

$T_1=400K;l_2=38cm;n_1=\alpha moles$

$50\%hgisspilledout\left(Wt\alpha VandV\alpha 1\right)$

$\therefore lenghtofHgleft=38cm\therefore lengthofgascoloum=152-38=114am$

$P_2=76+38=114;l_2=144cm$