If the value of `C_(p)` for nitrogen gas is 7 `JK^(-1)mol^(-1)`, then the value of `DeltaH` on heating 28 g of nitrogen gas from `0^(@)C` to `100^(@)C` at constant pressure will be

Options

A. 1200 J
B. 700 J
C. 1400 J
D. 1500 J

Answer

B

Explanation

Given, that, `C_(p)=7KJ^(-1)mol^(-1)`
weight of `N_(2)` gas=28g
`T_(1)=0^(@)C=273+0=273K`
`T_(2)=100^(@)C=273+100=373K`
Number of moles `=("weight of "N_(2))/("molecular weight of "N_(2))=(28)/(14)=2`
Now, by using `DeltaH=nC_(p)(T_(2)-T_(1))=2xx7(373-273)`
`=14xx100=1400J`