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pls explain   the work done in   isothermal reversible expansion  of an ideal gas

Asked by ppratim02 25th January 2016, 6:16 PM
Answered by Expert

Consider an ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston. The cylinder is not insulated. The external pressure, Pext is equal to pressure of the gas, Pgas.

 Pgas  = Pgas  = P

 If the external pressure is decreased by an infinitesimal amount dP, the gas will expand by an infinitesimal volume, dV. As a result of expansion, the pressure of the gas within the cylinder falls to Pgas – dP, i.e., it becomes again equal to the external pressure and, thus, the piston comes to rest. Such a process is repeated for a number of times, i.e., in each step the gas expands by a volume dV.

 Since the system is in thermal equilibrium with the surroundings, the infinitesimally small cooling produced due to expansion is balanced by the absorption of heat from the surroundings and the temperature remains constant throughout the expansion.

The work done by the gas in each step of expansion can be given as,
dw = –(Pext – dP) dV = –Pext.dV – dP.dV
The product of two infinitesimal quantities, is negligible.
The total amount of work done by the isothermal reversible expansion of the ideal gas from volume V1 to volume V2 is, given as,
w = –nRT loge V2/V1 or w = –2.303nRT log10 V2/V1
At constant temperature, according to Boyle’s law,
P1V1 = P2V2 or V2/V1 = P1/P2 So, w = –2.303nRTlog10 P1/P2
Answered by Expert 25th January 2016, 6:21 PM
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