# Mention the condition when qp = qv final internal change for the reaction A(l) ----> A(g) at 373 k heat of vaporisation is 40.66 kj R = 8.314 J+mole+k ?

Dear student

For the first question,
Qp is the heat change taking place at constant pressure and Qv is the heat change taking place at constant volume,
Both of them are related as:
${q}_{P}={q}_{v}×\left(RT{\right)}^{\Delta {n}_{g}}$
If the change in the number of moles of gaseous reactants is 0 for a reaction, then the value of Qp and Qv will be the same.

Now for the second question we are given that
$A\left(l\right)\to A\left(g\right)\phantom{\rule{0ex}{0ex}}a+1=40·66kJ\phantom{\rule{0ex}{0ex}}T=373k$
Since qp is delta H and qi is delta U
so from the given data, one can say that
${q}_{P}=\Delta H=40.66×1000\phantom{\rule{0ex}{0ex}}{q}_{v}=\Delta u\phantom{\rule{0ex}{0ex}}{q}_{P}={q}_{v\left(RT{\right)}^{\Delta ng}}$
$40660=\Delta \cup {\left(8.314×373\right)}^{{n}_{g}}$
$\Delta {n}_{g}=1-0=11\phantom{\rule{0ex}{0ex}}40660=\Delta u\left(8·314×373\right)\phantom{\rule{0ex}{0ex}}\Delta U=13·11J$

Regards

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