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Thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThermodynamics5 Marks
Question
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3J 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 = 4.19J)
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Answer

The work done (W) on the system while the gas changes from state A to state B is 22.3J. This is an adiabatic process. Hence, change in heat is zero.$\therefore\Delta\text{Q} = 0$
$\Delta\text{W} = -22.3\text{ J}$ (Since the work is done on the system)
From the first law of thermodynamics, we have: $\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$ Where, ΔU = Change in the internal energy of the gas$\therefore\ \Delta\text{U}=\Delta\text{Q}-\Delta\text{W}=-(-22.3\text{ J})$
$\Delta\text{U} = +22.3\text{ J}$
When the gas goes from state A to state B via a process, the net heat absorbed by the system is:$\Delta\text{Q} = 9.35\text{ cal}$
= 9.35 × 4.19 = 39.1765J Heat absorbed, $\Delta\text{Q}=\Delta\text{U}+\Delta\text{Q}$$\therefore\ \Delta\text{W}=\Delta\text{Q}-\Delta\text{U}$
= 39.1765 - 22.3 = 16.8765 Therefore, 16.88J of work is done by the system.

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