A gas present in a cylinder fitted with a frictionless pistion expands against a constant pressure of `1atm` form a volume of `2L` to a volume of `6L`. In doing so, it absorbs `800J` heat form the surroundings. Determine the increases in internal enegry of process.

Explanation

Since, work is done against constant pressure and thus, irreversible.
Given, `DeltaV = (6-2) = 4L`,
`P =1 atm`
`W =- 1 xx 4L-atm = -(1xx4 xx 1.987)/(0.0821)cal`
(Since `0.0821 L-atm = 1.987 cal)`
`=- 96.81 cal =- 96.81 xx 4.184 J ( :'1 cal = 4.184J)`
`=- 405.05 J`
Now, from the first law of thermodynamics
`q = DeltaU - w`
`800 = DeltaU +405.05`
`:. DeltaU = 394.95J`