Two moles of an ideal gas were expanded isothermally against opposing pressure of 1 atm from 20L to 60L. Compoute w, q, `DeltaE and DeltaH` for the process is joule. (Given `1L*atm=101.3J`)

Explanation

We know, `w=-P_(ex)(V_(2)-V_(1))`
`therefore w=-1(60-20)=-40L*atm=-40xx101.3J`
`=-4.052kJ`
For this process `DeltaE=0 and DeltaH=0` because the process is isothermal and the system is ann ideal gas. As per the first law of thermodynamics, `DeltaE=q+w`
or, `0=q-4.052kJ`
`therefore q=4.052kJ`