One mole of an ideal monoatomic gas `(C_(V.M)= 1.5 R)` is subjected to the following sequence of steps : (a) The gas is heated reversibly at constant

Monjur Alahi

One mole of an ideal monoatomic gas `(C_(V.M)= 1.5 R)` is subjected to the following sequence of steps :
(a) The gas is heated reversibly at constant pressure of 1 atm from 298 K to 373 K.
(b) Next, the gas is heated reversibly and isothermally to double its volume.
(c) Finally , the gas is cooled reversibly and adiabatically to 308 K .
Calcuated q, w, `DeltaH` for the overall porcess.


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Salim Ahmed Kawsar

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Correct Answer - (a) `q= DeltaH = 1558.88, DeltaU = 935.33 ; W= - P(DeltaU) = - 623.55 J mol^(-1)`
(b) `W =- 2149.7 J//mol ; DeltaU & DeltaH = 0, q = - W`
(c) `q = 0, W =- 810.62 , DeltaH = - 1351.03J mol^(-1)`
for overall process `q = 3705.59 ; W = - 3583.88, DeltaU = 124.71 ; DeltaH = 207.85`
(a) `" "W= intP_("ext").dV = - P_("ext). DeltaV`
`= - nRDeltaT`
`=- 1xx 8.14 xx (373 - 298)`
` = -623.55 J//mol`
`DeltaH = q = nC_(P)DeltaT`
`= - 2.5 xx 8.314 xx 75" " (C_(P)=2.5 R)`
`= 1558.8 J//mol`
`DeltaU = 1.5 xx 8.314 xx 75" "(C_(V) = 1.5 R)`
` =935.5 J//mol`
(b) ` W =-2.303RT log ((V_(1))/(V_(2)))`
` = - 2.303 xx 8.314 xx 373 log 2`
` = - 2149. 7 J//mol`
`DeltaU = 0 , DeltaH = 0 " "q =2149.7 J//mol^(-1)`
(c) `W = Deltau = nC_(V)DeltaT" "("For adiabatic process "q = 0, DeltaU ne W)`
`= 1 xx 1.5 xx 8.314 (308-373)`
` = - 810.62 J//mol`
`q= 0`
`DeltaH = nC_(P)DeltaT =- 1 xx 2.5 xx 8.314 (308-373)`
`= - 1351 .03 J//mol`
For overall process `q_(net) = 3708.59 J//mol`
`W_("net") = - 3583. 88 J//mol" "DeltaU_("net")= 124.71 J//mol`
`DeltaH_("net")= 207.85 J//mol`

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