In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 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)
Given, the work done on the system is 22.3 J , heat absorbed by the system is 9.35 J and 1 cal is equivalent to 4.19 J .
Let Δ Q be the change in heat, then
Δ Q = 0
The above result is correct because the given process is an adiabatic process.
Let Δ W be the work transfer from the system, so
Δ W = − 22.3 J
Negative sign indicates that work is done on the system.
Let Δ U be the change in the internal energy of the system. So, from the first law of thermodynamics,
Δ Q = Δ U + Δ W (1)
Substitute the values in the above expression,
0 = Δ U + ( − 22.3 ) Δ U = 22.3 J
When the system goes from state A to state B by absorbing 9.35 cal of heat, then
Δ Q ′ = 9.35 cal =9 .35 × 4.19 = 39.176 J
Use the first law of thermodynamics from equation (1) and substitute the values.
Δ Q ′ = Δ U + Δ W ′ Δ W ′ = 39.1765 − 22.3 ≈ 16.9 J
Hence, the work done by the system is 16.9 J .
Given, the work done on the system is
Let
The above result is correct because the given process is an adiabatic process.
Let
Negative sign indicates that work is done on the system.
Let
Substitute the values in the above expression,
When the system goes from state
Use the first law of thermodynamics from equation (1) and substitute the values.
Hence, the work done by the system is
