For the reaction, `2 CO+ O_(2) to 2CO_(2), Delta H= - 560kJ`. 2 mol CO and 1 mol of `O_(2)` are taken in container of volume 1L. They completely form two moles of `CO_(2)`. The gases deviate apreciably form ideal behaviour. If the pressure in the vessel changes from 70 to 40 atm, find the magnitude (absolute value) of `Delta U` at 500 K.

Explanation

`H= U+ PV therefore Delta H = Delta (U+PV)= Delta U+Delta (PV)`
At constant volume, `Delta H= Delta U+V Delta P = Delta U + V(P_(2)-P_(1))`
Given : `Delta H= - 560 kJ , P_(1)= 70 " atm", P_(2)= 40 " atm &" V= 1L`
`therefore - 560 kJ = Delta U+ 1(40-70)= Delta U- 30 " L atm "= Delta U- 3kJ`
`therefore Delta U= - 557 kJ`.
Hence, `|Delta U|= 557 kJ`.