1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature, absorbs heat 3710 J and expands to 2 litres. Calculate the entropy change in expansion process.

Explanation

Given : n=1 mole
P=4.1 atm
V=2 Lit
T=?
q=3710 J
`DeltaS=q/T`
`DeltaS=q/((PV)/(nR))`
`DeltaS=(nRq)/(PV)`
`DeltaS=(1xx0.082 "lit atm" K^(-1) xx3710 J)/("4.1 atm" xx "2 lit")`
`DeltaS=37.10 JK^(-1)`