An insulated vessel contains 1mole of a liquid, molar volume `100mL` at 1bar. When liquid is steeply passed to `100`bar, volume decreases to `99mL`. Find `DeltaH` and `DeltaU` for the process.

Answer

`Delta U = 100` bar ml; `Delta H = 9900` bar ml

Explanation

Here, `P_(1) = 1` bar `P_(2) = 100` bar, `V_(1) = 100 ml, V_(2) = 99 ml`
For adiabatic process, `q = 0 , Delta U = W`
Since, `Delta U = q + W` (first law of thermodynamics)
`=q - P (V_(2) - V_(1)) [W = P (V_(2) - V_(1))`
`= 0 - 100 (99 - 100) = 100` bar mL
Also, `Delta H = Delta U + Delta (PV) = Delta U + P_(2)V_(2) - P_(1) V_(1), = 100 + (99 xx 100 - 100 xx 1) = 9900` bar mL