Question

6. (25 points) One mole of a monatomic ideal gas, initially at pressure P1 = 105 Pa and temperature T1 = 273 K undergoes an isovolumetric process in which its pressure falls to half its initial value. a) What is the work done by the gas? What is the final temperature? b) The gas then expands isobarically (constant pressure) to twice its initial volume. What is the work done by the gas? What is the final temperature? c) Draw a PV diagram of the two step process.

Answer 1

6. One more of ichal gas. Initial pressure is P,= 105 Pe, indtial temprature T, =273K the gas goes isovelimetric fisocess and the final pressure P = 105. (a) The work done by the gas is was Por zo because there's no volume change since, the gas is monoatomic ideal, it follows the ideal gas law: P,V, = RTI P2 V₂ a RT But, Vi = V₂ so, - T o The entity o Tz = ( x 273 o a 273 ALL 2 (b) the gas expands isobarically now, so fresslese is - constant and the final volume be comes twice 3 the ênitial volume. - so i Va=zv. )

Pdv But Pa RT W = RTL r = RT in Yll « Wo RT line video . put V2 = 24 - C W = RT ln 2 R= 8.3 14 T/mol-k ༡།? །ཏེ །? ? ། ། །? ག ད ག ག ན། ན) ན། ན་ su Wa 8.314x273 x Inz T W = 1573. 25 J) cimsuses Now the final temprature is obtained by Pri= RTI P2 V2 2 RT2 :P,=P so, y = I so, Tz = T. ve 1 Ta= 2x273 k)

processes in (c) The diagrams for for the the lo two pa (los) e f (PTM (P, 72, V2) --(PUTOV processes of ☺ Cises @ 10. as shown above represent