Chemistry - Thermodynamics - II Question with Solution | TestHub

Reversible isothermal expansion from$\mathrm{I}\to \mathrm{II}$during this process

$\mathrm{T}=$ constant

$\mathrm{\Delta H}=\mathrm{\Delta U}=0$

$\mathrm{H}=$ constant.

$\mathrm{\Delta S}\ne 0$ Entropy will increase $\mathrm{PV}=$ constant.

and during adiabatic reversible expansion.

$\mathrm{q}=0$

$\mathrm{\Delta U}=\mathrm{w}=-\mathrm{ve}$

so $\mathrm{T}$ will decrease

$\mathrm{\Delta H}={\mathrm{nc}}_{\mathrm{p}}\mathrm{\Delta T}=-\mathrm{ve}$

$\mathrm{H}$ will decrease

$\mathrm{\Delta S}=0$ (isoentropic process)

$\mathrm{S}=$constant. and ${\mathrm{pv}}^{\mathrm{r}}=$ constant.

Keeping in mind correct graph are 1,2 and 4.