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q = 0 as adiabatic process is given

            W = 0 as p_axt = 0

            q + W = DU

            q = 0

            W = 0

            Þ DU = 0

In air $F=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r^2}$  

In medium $F\text{'}=\frac{1}{4\pi \left(k{\in }_0\right)}\frac{q_1{q}_2}{r^2}$  

$F\text{'}=\frac{F_0}{K}$

PF₅, PCl₅, PBr₅, Fe (CO)₅ Þ Trigonal bipyramidal

BrF₅ Þ Square pyramidal

[PtCl₄]^2– Þ Square planar

SF₆ Þ Octahedral

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$a=\frac{\left(m_1-m_2\right)g}{\left(m_1+m_2\right)}=\frac{g}{8}$

8m₁ – 8m₂ = m₁ + m₂

7m₁ = 9m₂

$\frac{m_1}{m_2}=\frac{9}{7}$          

U = nC_vT

-> $U=n_1{C}_{v_1}T+n_2{C}_{v_2}T$  

-> $8*\frac{5R}{2}*T+4*\frac{5R}{2}*T$

= 30RT

Boiling point: $\stackrel{\left(239.7\right)}{N{H}_3}>\stackrel{\left(185.5\right)}{P{H}_3}$  due to hydrogen bonding in NH₃.