10 mole of an ideal gas is heated at constant pressure of one atmosphere from `27^(@)C` to `127^(@)C`. If `C_(v,m)=21.686+10^(-3)T(JK^(-1).mol^(-1))`, then `DeltaH` for the process is :

Options

A. 3000 J
B. 3350 J
C. 3700 J
D. `3181.4J, 2350J`

Answer

D

Explanation

`C_(p,m)=(21.686+8.314+10^(-3)T)`
`=30+10^(-3)T`
`intDeltaH=intnC_(p,m)DeltaT`
`=10xxint(30+10^(-3)T).dT`
`DeltaH=10xx30(T_(2)-T_(1))+(10^(-3))/(2)xx[T_(2)^(2)-T_(1)^(2)]xx10`
= 30350 J