Question

12. 1 mole of an ideal gas undergoes an isothermal expansion from V1 = 1.4L followed by isobaric compression, p = cst.if P1 = 4.4atm, p2 = 1.7atm → ?- m calculate the work done by gas during the expansion. Express work in J = N·m! • For isothermal processes, AT = 0 T = cst → w=faw=fr&v=/MRT AV 594 Show your work like: `x-int_0^5 v(t)dt rarr x-int_0^5(-4*t)dt=-50 m 13. 1 mole of an ideal gas undergoes an isothermal expansion from V1 = 1.7L followed by isobaric compression, p = cst.if P1 = 3.2 atm, P2 = 1.4atm, calculate N the total work done by gas during the compression → ? m2 • The total work done by the gas is the sum of the works done for the two processes, isothermal and isobaric. For isothermal processes, AT = 0+1 = cst → Hint: W= = aw = [ pav V= / "RT AV Hint 1: Constant Pressure Process Hint 2: Constant Volume Process Hint 3: Isothermal Process Hint 4: Adiabatic Process Hint 5: Heat Engine Cycle Hint 6: Heat Engine Cycle

Answer 1

(12) Р A (10 42,4 4 atm) 1. Fatm v Temperature at a PV=RT 5 T PV 1.4x10²x 4.4x10 к 8.314 T = 74.1k The temperature is same every where on iso therm, re. the temperature at B = 74.1k Volume of the gas at B Pr= RT V = RT P 8.31 X 74.1 1o Fox 105 3.62 L VE 2. The work done by the gas W = SORT av Vi

W = NRT Ve loge = 1x 8-31*74.1 x log (9:40) 585 N.m (13) P A (1.76, 3.2 afm) B 1-4 atm Tempe rafure at a from Ideal gas equation PV=nRT PV NR 1- Ixió x 3-2x16 1 x 8.314 - 65.5k AB is an is otherm, So temperature will be the same at is volume of the gars at B v= nRT P 1x8.31865-5 1.4 x 105 = 3.89 L

work done in isothermal process w = v ner av = nRT loge (*) 3.89 = 1x8-31X65-5x logo = 451 Nm work done in isobaric compression RA p (v.-v2) = 1.4x10 3 16 (1-4 - 3-89) - 307 N.m Net work done W = Wet Wp = 451 +- 307 = 144 Nm