One mole of an ideal mono atomic gas is at a temperature $T_{0} = 1000 K$ and its pressure is $P_{0}$ The gas is adiabatically cooled so that its pressure becomes $\frac{2}{3} P_{0}$ .Thereafter, the gas is cooled at constant volume to reduce its pressure to $\frac{P_{0}}{3}$ . The total heat absorbed by the gas during process in joules is Q.
Write value of $\frac{4 * Q}{425}$
Given: $R = \frac{25}{3} J m o l^{- 1} K^{- 1} a n d {(\frac{2}{3})}^{2 / 5} = 0.85$

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Detailed Solution

For the adiabatic process, we have
$P_{2}^{1 - T} T_{2}^{T} = P_{1}^{1 - T} T_{1}^{T}$
$\Rightarrow \frac{T_{2}}{T_{1}} = {(\frac{P_{1}}{P_{2}})}^{\frac{1 - T}{T}}$
$T_{2} = 1000 {(\frac{3}{2})}^{\frac{3}{5} - 1} = 1000 {(\frac{2}{3})}^{2 / 5} = 1000 \times 0.85 = 850 K$
For the isochoric process
$\frac{P_{3}}{T_{3}} = \frac{P_{2}}{T_{2}} \Rightarrow T_{3} = (\frac{P_{3}}{P_{2}}) T_{2}$
$\Rightarrow T_{3} = (\frac{P_{0} / 3}{2 P_{0} / 3}) \times 850$
$\Rightarrow T_{3} = \frac{850}{2} = 425 K$
Heat is lost in 2nd process only
Heat lost= $= n C v Δ T = 1 \times \frac{3}{2} \times R \times (850 - 425)$
$= \frac{3}{2} \times \frac{25}{3} \times 425 = 5312.5 J$