According to law of conservation of energy
Total energy = work done + internal energy
ΔQ = ΔW + ΔU
Rate at which heat is supplied ΔQ = 100 W
Rate at which work is done ΔW = 75 Js-1
Rate of change of internal energy is ΔU
ΔU = ΔQ – ΔW
ΔU = 100 – 75
ΔU = 25 J s-1
The internal energy of the system is increasing at a rate of 25 WAnswered by Pragya Singh | 1 year ago
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Figure.
Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?
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)