`N_(2)` gas is heated from 300 kg temperature to 600 k through an isobaric process. Then find the change in the entropy of the gas (n = 1 mole )

Options

A. 10 J/k
B. 20 J/k
C. 30 J/k
D. 40 J/k

Answer

B

Explanation

`d theta = nC_(p) d T`
` ds =( d theta)/(T) =(nC_(p)dT)/(T) `
`Delta S=nC_(p) overset(T_(2))underset(T_(1))int (dT)/(T) =nC_(p)"In" ((T_(2))/(T_(1))) `
`DeltaS=1 xx (7)/(2) R " In " ((600)/(300))=(7)/(2) xx 8.3 xx 0.693`
`DeltaS=20J//k`