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A sample of argon gas at 1 atm pressure and 27^(@)C expands reversibly and adiabatically from 1.25dm^(3) to 2.50 dm^(3). Calculatethe enthalpy changein this process. C_(v,m) for argonis 12.48 JK^(-1) mol^(-1)

Answer» <html><body><p></p>Solution :For adiabaticprocess , <br/> `T_(1)V_(1)^(gamma-1)= T_(2)V_(2)^(gamma-1) ` or `((V_(1))/(V_(2)))^(gamma-1) = (T_(2))/(T_(1))` <br/> But `gamma=( C_(p))/(C_(<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>))` <br/> `:gt gamma -1 = (C_(p))/(C_(v))-1= ( C_(p)-C_(v))/(C_(v))= (R)/(C_(v))` ,brgt <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> `((V_(1))/(V_(2)))^(R//C_(v))= (T_(2))/(T_(1))` <br/> or `ln.(T_(2))/(T_(1))=(R)/(C_(v))ln.(V_(1))/(V_(2)) ` or `log.(T_(2))/(T_(1))= (R)/(C_(v)) log. (V_(1))/(V_(2))` <br/> i.e.,`log. (T_(2))/(300) = ( 8.314)/( 12.48) log. ( 12.5)/(25)` <br/> or ` logT_(2) = log 300 + 0.666 log . ( 1)/(2)` <br/> `= 2.4771 + 0.666 ( - 0.3010) = 2.2766` <br/> or`T_(2) =` Antilog 2.2766 `=<a href="https://interviewquestions.tuteehub.com/tag/189-1799650" style="font-weight:bold;" target="_blank" title="Click to know more about 189">189</a> K`<br/> `C_(p)=(DeltaH)/( DeltaT)`<br/>or `DeltaH =C_(p) Delta T ` or for <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> moles `Delta H = n C_(p) Delta T`.<br/> Taking the gas as ideal. <br/> `PV = nRT ` or ` 1 xx 1.25 = n xx 0.0821 xx 300` or `n = 0.05` mole <br/> Further, `C_(p) = C_(v) + R = 12.48 + 8.314 = 20.794J` <br/> `:.Delta H = 0.05 xx 20.794 xx ( 189 - 300) = - 115 . 4 J`</body></html>


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