InterviewSolution
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InterviewSolution
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If one mole of an ideal gas with `C_(v) = (3)/(2) R` is heated at a constant pressure of 1 atm `25^(@)C` to `100^(@)C`. Which is correctA. `Delta U` during the process is 223.51 calB. `Delta H` during the process is 372.56 calC. entropy change during the process is `1.122 cal k^(-1) mol^(-1)`D. `Delta U, Delta H` are same for the process |
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Answer» Correct Answer - A::B::C `because C_(v) = (3)/(2)R :. C_(p)=C_(v) + R = (3)/(2)R+R = (5)/(2)R` `:.` Heat given at constant pressure ` m.C_(p).Delta T` ltbr. Or Now work done in the process `= -P Delta V` `Delta H` or `q_(p) = 1 xx (5)/(2) xx R xx (373-298)` or `Delta H = 1 xx (5)/(2) xx 1.987 xx 75 = 372.56 cal` `w = -p(V_(2)-V_(1)) = -p((nRT_(2))/(p)-(nRT_(1))/(p))` `(because pv = nRT)` `= -nRT(T_(2)-T_(1)) = -1 xx 1.987 xx (373 - 298)` `= -149.225 cal` `:.`from I law of thermodynamics `Delta U = q+Q = 372.56 - 149.05` `:. Delta U = 223.51 cal` Also, `dq_(rev) = nC_(p).dt` `ds = (dq_(rev))/(T) :. ds =(nC_(p).dt)/(T)` or `Delta S = int_(T_(1))^(T_(2))(nC_(p).dT)/(T) = nC_(p)log_(e).(T_(2))/(T_(1))` `:. Delta s = 2.303 nC_(p) log_(10).(T_(2))/(T_(1))` `= 2.303 xx 1 xx (5)/(2) xx (5)/(2) R xx log_(10).(373)/(298) =1.122 cal k^(-1) mol^(-1)` |
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