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?
(a) Now, as soon as the stop cock is opened the gas will start flowing from cylinder P to cylinder Q which is completely evacuated and thus the volume of the gas will be doubled because both the cylinders have equal capacity. And since the pressure is inversely proportional to volume, hence the pressure will get decreased to half of the original value.
Since, the initial pressure of the gas in cylinder P is 1 atm. Therefore, the pressure in each of the cylinder will now be 0.5 atm.
(b). Here, in this case, the internal energy of the gas will not change i.e. ΔU = 0. It is because the internal energy can change only when the work is done by the system or on the system. Since in this case, no work is done by the gas or on the gas.
Therefore, the internal energy of the gas will not change.
(c) There will be no change in the temperature of the gas. It is because during the expansion of gas there is no work being done by the gas.
Therefore, there will be no change in the temperature of the gas in this process.
(d). The above case is the clear case of free expansion and free expansion is rapid and it cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non – equilibrium states, they do not lie on the Pressure-Volume – Temperature surface of the systemAnswered 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
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)