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)
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.
∴ ΔQ = 0
ΔW = -22.3 J (Since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = ΔU + ΔW
Where,
ΔU = Change in the internal energy of the gas
∴ ΔU = ΔQ - ΔW = - (- 22.3 J)
ΔU = + 22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is:
ΔQ = 9.35 cal = 9.35 x 4.19 = 39.1765 J
Heat absorbed, ΔQ = ΔU + ΔQ
∴ΔW = ΔQ - ΔU = 39.1765 - 22.3 = 16.8765 J
Therefore, 16.88 J of work is done by the system.
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(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
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(b) the magnitude of average velocity? Are the two equal?
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(b) the force on the 7th coin by the eighth coin,
(c) the reaction of the 6th coin on the 7th coin.