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 \[22.3{\text{ }}J\] 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 \[9.35{\text{ }}cal\], how much is the net work done by the system in the latter case? (Take \[1\] cal =\[4.19{\text{ }}J\])

Answer 1

Hint: If there is no exchange of heat from the system to its surroundings during both compression and expansion in a thermodynamic process is called an adiabatic process. The adiabatic process can be both reversible and irreversible. In order to answer this question we need to find the change in the internal energy in the first case and then substitute it in the next case to find the work done:

Complete step by step solution:
First case: The state of the gas is adiabatically changed from an equilibrium state A to state B. While doing so the amount of work is equal to \[22.3{\text{ }}J\].
Since this work is done on the system the net work done will be negative. Therefore,
\[\Delta W = - 22.3J\]
We know that in an adiabatic process the net change in the heat is equal to zero.
\[\Delta Q = 0\]
 And the formula is given as,
\[\Delta Q = \Delta U + \Delta W\] …….. (1)
Here, \[\Delta U\]is said to be the change in the internal energy
\[\Delta W\]is said to be the change in the work.
Therefore substituting for \[\Delta W = - 22.3J\] in the above equation we get,
\[0 = \Delta U - 22.3J\]
\[\Delta U = 22.3J\].
Therefore the above equation gives the change in the internal energy.
In the latter process, both the initial and the final states are the same as those in the previous case therefore \[\Delta U\] remain the same for the latter case.
Second case: In this case, the state of the gas is taken from A to B. While doing so the net heat absorbed by the system is \[9.35{\text{ }}cal\]. Given that \[1\] calories \[ = 4.19{\text{ }}J\]. Therefore the net change in the heat is equal to,
\[\Delta Q = 9.35 \times 4.19 = 39.1765J\]
Substituting \[\Delta Q\] and \[\Delta U\] in equation (1) we get,
\[39.1765J = 22.3J + \Delta W\]
\[W = 39.1765J - 22.3J\]
\[W = 16.876J\]
Therefore the net work done in the latter case the amount of net work done is given as \[W = 16.876J\]

Note:
If a gas expands or compresses adiabatically the work done can be both positive and negative. This is assigned by two conventions to determine the sign of work. If the gas expands it means that the work is done on the gas and therefore the work done will be negative as we have seen in our case. If the gas is compressed it means that the work done by the system and therefore the work done will be positive.