Calculate `q,w,DeltaU`, and `DeltaH` for this isothermal reversible expansion of 1mole of an ideal gas from an initial pressure of `10.0` bar to final pressure of `100`bar at a constant pressure of `273K`.

Explanation

In isothermal process as temperature remains constant both `DeltaU` and `DeltaH` are zero.
`DeltaU = 0, DeltaH =0`
Applying the first law of thermodynamics.
`DeltaU = w +Q`
`0 = w +Q`
or `Q =- w =- 2.303 nRT log_(10). ((P_(1))/(P_(2)))`
`=- 2.303 xx1xx 8.314 xx273 log_(10). ((10)/(1000))`
`=5227.169 J = 5.227 kJ`