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Q. A gaseous system undergoes a change of state from (A) to (B) by any of the given path
path I or path II as shown in figure
As per path I;
$\Delta q=-400 \,cal$ and $\Delta W=14 \,cal$
As per path II, $\Delta q=-48\, cal$
image
Therefore work done, $\Delta W$ in path (II) is

Thermodynamics

Solution:

For path I $ \Delta E_{ I }=\Delta q+\Delta W=-400+14=-386\, cals$

For path II, $ \Delta E_{ II }=\Delta E_{ I }=-48+\Delta W$

$\therefore \Delta W=-386+48=-338\, cal$