A cylinder of circular cross-section and a 50-kg pistoncontain 0.120 mol of comp

Question

A cylinder of circular cross-section and a 50-kg pistoncontain 0.120 mol of compressed air, as depicted in the diagram.the piston is in mechanical equilibrium, and although the piston isable to maintain a seal to prevent leakage of the compressed air,friction between the cylinder wall and the piston is negligible.Standard gravity acts vertically downward. Standard atmosphericpressure is present. The initial temperature of the compressed airis 30 degrees Celcius. b) The compressed air is then slowly heated and therebyreaches a new equilibrium state for which the temperature change is100 degrees Celcius. Determine the change in the piston height fromthe previous state [in Part (a)]. c) The compressed air is then maintained at this newtemperature, but a block whose mass is identical to the piston isgradually lowered onto the top of the original piston. Determinethe change in the piston height from the previous state in [Part(b)].

Explanation / Answer

area=/4*0.12=7.85*10-3 m2 (a) at equi;ibrium position pressure*area=atmostpheric pressure+mg/area mg is weight of the piston n*R*T/V=105+50*9.8/area =0.162*106Pa n=0.12 mol R=8.31 T=30+273=303 K => V =h*area........=> h=0.238 m=23.8 cm (b)The compressed air is then slowly heatedand thereby reaches a new equilibrium state for which thetemperature change is 100 degrees Celcius (b)The compressed air is then slowly heatedand thereby reaches a new equilibrium state for which thetemperature change is 100 degrees Celcius the pressure of the air isconstant => V/T=constant => h/T=constant => h'=h*(100+273)/303=0.293 m=29.3 cm (c) the change in the piston height from the previousstate in [Part (b)]. PV=constant initial: P=0.162*106 Pa V=0.293*area final: P'=105+2*50*9.8/area =0.224*106 Pa V=area*x => x=P/P'*0.293=0.212 m change:0.293-0.212=0.081=8.1 cm (c) the change in the piston height from the previousstate in [Part (b)]. PV=constant initial: P=0.162*106 Pa V=0.293*area final: P'=105+2*50*9.8/area =0.224*106 Pa V=area*x => x=P/P'*0.293=0.212 m change:0.293-0.212=0.081=8.1 cm

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