If 20 gm `N_2` at 300 K is compressed reversibly and adiabatically from `20 dm^(3)` to 10 `dm^(3)`
Given: `[(2)^(2/5) = 1.32]`

Answer

2

Explanation

For reversible adiabatic process
`TV^(r-1)=K`
Moles of gas `N_2=20/28`
`N_2` is a diatomic gas
`C_(um)=5/2R`
`C_(pm)=7/2 R`
`r=7/5`
`T_1V_1^(r-1)=T_2V_2^(r-1)`
`T_2/T_1=(upsilon_1/upsilon_2)^(r-1)`
`T_2/300=(20/10)^(7/(5^(-1)))=(2)^(2/5)=1.32`
`impliesT_2=396K`
For adiabatic
`DeltaV=w_(rev)=nc_(um)(T_2-T_1)`
`DeltaH=nC_(p)m(T_2-T_1)`
`=20/28 xx 7/2 xx R xx 98 = 240 xx 25/3`
`=2000 J`
`DeltaH=2KJ`