Question

ProcessIn changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal $=$ 4.19 J)

Answer 1

The work done (W) on the system while the gas changes from state A to state B is 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.

$\therefore △Q=0$

$△W=-22.3J$ (Since the work is done on the system)
From the first law of thermodynamics, we have:

$△Q=△U+△W$
Where,
$△U=$ Change in the internal energy of the gas

$\therefore △U=△Q-△W=-(-22.3J)$

$△U=+22.3J$
When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

$△Q=9.35cal=9.35\times 4.19=39.1765J$
Heat absorbed, $△Q=△U+△W$

$\therefore W=△Q-△U$
$=39.1765–22.3$
$=16.8765J$
Therefore, 16.88 J of work is done by the system.