MCQ
$\frac{1}{24.63}$ mole of an ideal mono-atomic gas undergoes a reversible process for which $PV^3= C$ where $C (L^3 -$ $atm$) is a constant. The gas expanded from initial volume of $1\, L$ and initial temperature of $300K$ to final volume of $2\, L$ then heat lost to the surroundings by gas is ......$J$
  • A
    $15.2$
  • $75.9$
  • C
    $37.9$
  • D
    $22.8$

Answer

Correct option: B.
$75.9$
b
$\mathrm{C}=1 \quad(\mathrm{P}=1, \mathrm{V}=1, \text { initially })$

$\mathrm{w}=-\int_{1}^{2} \mathrm{PdV}=-\int_{1}^{2} \frac{1}{\mathrm{V}^{n}} \mathrm{d} \mathrm{V} \Rightarrow$

$-\left[\frac{V^{-n+1}}{-n+1}\right]_{1}^{2}=\frac{1}{n-1}\left[\frac{1}{2^{n-1}}-1\right]=-0.375\, \mathrm{L}-\mathrm{atm}$

$\mathrm{V}^{2} \mathrm{T}=\mathrm{C}$

$\mathrm{T}_{\ell}=75\, \mathrm{K}$

$\Delta E=\frac{1}{24.63} \times \frac{3}{2} \times 0.0821[75-300]$

$=-1.125\, \mathrm{L} \text { -atm }$

$\Delta \mathrm{E}=\mathrm{q}+\mathrm{w}$

$=-1.125\, \mathrm{L}-\mathrm{atm}=-0.375+\mathrm{q}$

$\Rightarrow \mathrm{q}=-0.75\, \mathrm{L}-\mathrm{atm}=75.9\, \mathrm{J}$

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