Chemistry - Thermodynamics - I Question with Solution | TestHub

$\mathrm{AB}\to$ isobaric

$\mathrm{BC}\to$ isochoric

$\mathrm{CA}\to$ not defined

For cyclic process, ΔU=0

$∆{\mathrm{U}}_{\mathrm{AB}}=\mathrm{q}+\mathrm{W}$

$=2-5=-3$

$∆{\mathrm{U}}_{\mathrm{ABC}}=∆{\mathrm{U}}_{\mathrm{AB}}+∆{\mathrm{U}}_{\mathrm{BC}}$

$=-3-5=-8\mathrm{ }\mathrm{KJ}$

$∆{\mathrm{U}}_{\mathrm{CA}}=+8$

$=\mathrm{Q}+\mathrm{ }\mathrm{W}$

$8=\mathrm{Q}+3$

$\mathrm{Q}=+5\mathrm{ }\mathrm{kJ}$

Therefore, the heat absorbed by the system during process CA is $+5 {\mathrm{KJmol}}^{-1}$.