MCQ
Which of the following process will give maximum amount of heat to surrounding when volume becomes half of initial
  • Iso-baric
  • B
    Isothermal
  • C
    Isochoric
  • D
    Adiabatic

Answer

Correct option: A.
Iso-baric
a
Here isochoric process is not possible and for

adiabatic process, $\mathrm{Q}=0$ For isothermal process

${Q_{rejected}}\, =  - W = nR{T_0}\ell n\left( {{V_0}/\frac{{{V_0}}}{2}} \right) = nR{T_0}\ell n\,2$

$=0.693 \mathrm{nRT}_{0}$

For isobaric process

$Q_{\text {rejected }}=-n C_{p} \Delta T=-n\left(\frac{f}{2}+1\right) n R\left(-\frac{T_{0}}{2}\right)$

$=\left(0.5+\frac{f}{4}\right) n R T_{0}>0.693 n R T_{0}$

It is clear that more heat is rejected in isobaric process.

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