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 correct

Options

A. `Delta U` during the process is 223.51 cal
B. `Delta H` during the process is 372.56 cal
C. entropy change during the process is `1.122 cal k^(-1) mol^(-1)`
D. `Delta U, Delta H` are same for the process

Answer

A, B, C

Explanation

`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)`