MCQ
One mole of Argon is heated using $PV^{5/2} =$ constant. By what amount heat is absorbed during the process, when temperature changes by $\Delta T = 26\ K$ ......$J$
  • A
    $100$
  • B
    $200$
  • $180$
  • D
    $208$

Answer

Correct option: C.
$180$
c
$\mathrm{n}=1 \,\,\,\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}$

$\mathrm{PV}^{5 / 2}=$ constant $, \mathrm{n}=5 / 2$

$\mathrm{q}=\mathrm{n} \mathrm{C} \Delta \mathrm{T}$

$ = {\text{n}}\left[ {{{\text{C}}_{\text{v}}} + \frac{{\text{R}}}{{1 - {{\text{n}}^\prime }}}} \right]\therefore \,\Delta {\text{T}}$

$=1\left[\frac{3}{2} R+\frac{R}{1-\frac{5}{2}}\right]$

$=\left[\frac{3}{2} R-\frac{2}{3} R\right] 26$

$=\frac{5}{6} \mathrm{R} \times 26=\frac{5 \times 8.314 \times 26}{6}=180\, \mathrm{J}$

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