Question

vf uf OR DO An ideal gas is very slowly expanded to ten times its original volume to atmospheric pressure (1.0 atm) such that the temperature is constant (T-2018°F) Calculate the initial pressure and final volume of the vessel as well as the work that was done by the system on the surroundings. Are these three calculated values intensive or extensive quantities? ii)Why is there a negative sign in front of the integral? What does it mean when the value for the work is positive/negative?

Answer 1

ven Final volume , Va is den timas to erigimal velume, V Thus ½=loy ) ,rns.tar责 土r_r Pt P. &드 Pi V 1.013a5 (substitute bol term equation 0) lov, - P 10 133 bas To jind initiol and kinal volume Fe an ideal gas, P- RT/v e PV RT Thus Vi = RT whose R→Gasc nstant V,-8.314 X12T6 48 v, = 0, 011& m3 0118 m3 Bystem lo.13&

For a comstant dempesatuaeprcess Vi where w → hJoek done Juj the sagtem V,->Initial volume vFinal velume B → initial Poesscuse Pi Soluing er w. 1.013

i) Please refer the attached image for calculations.

Pressure is an intensive property as it is independent of the size of the system.

Volume is an extensive property as it depends on the quantity of matter specified in the system.

Work involved in a given change of state are not determined only by the initial and final states, they also depend on the path in which the change is carried out. Hence Work is not thermodynamic property of the system, but are properties of the process and are called path functions.

ii) For expansion process V₂ > V₁ and ln (V₂/ V₁) is greater than zero. Since R and T are always positive, the work done by the ideal gas at constant temperature is always positive. Hence there is negative sign before integral.

Work is positive if work is done by the system. It is negative if work is done on the sytem.