For the gas \(C_V=20\,J/K.\) Thus, \(20\,J\) of heat is required for \(1^o\) change in temperature of the gas.
Heat change involved during the process
i.e., work done \(3\,kJ=3000\,J\)
Change in temperature \(=\frac {3000}{20}\,K=150\,K\)
Initial temperature \(=300\,K\)
Since, the gas expands so the temperature decreases and thus final temperature is
\(300-150=150\,K\)
$(i)$ $N_2H_4$$_{(l)}$ $+$ $2H_2O_2$$_{(l)}$ $\rightarrow$ $N_2$$_{(g)}$ $+$ $4H_2O$$_{(l)}$; $\Delta r{H_1}^ \circ = - 818 \,kJ/mol$
$(ii)$ $N_2H_4$$_{(l)}$ $+$ $O_2$$_{(g)}$ $\rightarrow$ $N_2$$_{(g)}$ $+$ $2H_2O$$_{(l)}$; $\Delta r{H_2}^ \circ = - 622 \,kJ/mol$
$(iii)$ ${H_2}_{(g)}\,\, $+$ \,\,\frac{1}{2}\,{O_2}_{(g)}\,\, \to \,\,{H_2}O_{(l)}\,\,\,;\,\,{\Delta }r{H_3}^ \circ \, = \,\, - 285\,\,kJ/mol$
$HCl + NaOH \rightarrow NaCl + H _{2} O \Delta H =-57.3\, kJ\,mol ^{-1}$
$CH _{3} COOH + NaOH \rightarrow CH _{3} COONa + H _{2} O$
$\Delta H =-55.3\,kJ\,mol ^{-1}$
વિદ્યાર્થી દ્વારા ગણતરી કરેલ $CH_3COOH$ની આયનિકરણ એન્થાલ્પી $......\,kJ\, mol ^{-1}$ છે.
$N{H_{3(g)}}\, + \,\,\frac{3}{2}\,Cu{O_{(s)}}\, \to \,\,\frac{1}{2}\,{N_{2(g)}}\, + \,\,\frac{3}{2}{H_2}{O_{(\ell )}}\, + \,\,\frac{3}{2}\,C{u_{(s)}}.$ ......$J$
( $R=2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ આપેલ છે.)
$\left[\right.$ ઉપયોગ $: {H}^{+}({aq})+{OH}^{-}({aq}) \rightarrow {H}_{2} {O}: \Delta_{{\gamma}} {H}=-57.1\, {k} {J} \,{mol}^{-1},$
વિશિષ્ટ ઊર્જા ${H}_{2} {O}=4.18 {Jk}^{-} {g}^{-},$
ઘનતા ${H}_{2} {O}=1.0\, {~g} {~cm}^{-3},$
મિશ્રણ પર દ્રાવણના કદમાં કોઈ ફેરફાર થતો નથી એમ ધારો.]