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A 20 litre container at 400 K contains CO_(2) (g)at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container . The maximum volume of the container when pressure of CO_(2) attains its maximum value, will be ("Given that " : SrCO_(3)(s) hArr SrO (s) + CO_(2) (g)K_(p) =1*6 atm)

Answer» <html><body><p>5 litre <br/>10 litre<br/>4 litre<br/>2 litre </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/maximum-556915" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMUM">MAXIMUM</a> pressure of `CO_(2)` = Pressure of `CO_(2)` at equilibrium <br/> For the reaction <br/> `SrCO_(3) (s) <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> <a href="https://interviewquestions.tuteehub.com/tag/sro-632549" style="font-weight:bold;" target="_blank" title="Click to know more about SRO">SRO</a> (s) + CO_(2) (g) `<br/> `K_(p) = p_(CO_(2))= 1.6 " <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a> = maximum pressure of "CO_(2)`<br/> Volume of container at this stage<br/> `V=nRT/P""` ...(i) <br/> <a href="https://interviewquestions.tuteehub.com/tag/intially-2132834" style="font-weight:bold;" target="_blank" title="Click to know more about INTIALLY">INTIALLY</a>, no . of moles of the gas (n) <br/>` (PV)/(RT)= (0.4 xx20)/(RT)= 8/(RT)` <br/> As no. of moles of the gas will remain constant , therefore, at equilibrium also, <br/> `n=8/(RT)` <br/> Putting this value in eqn. (i), <br/>`V= 8/(RT)*(RT)/P = 8/P= 8/1.6=5L`</body></html>


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