Consider a piston-cylinder assembly containing 10kg of water.Initially the gas h

Question

Consider a piston-cylinder assembly containing 10kg of water.Initially the gas has pressure of 20 bar and occupies a volume of1.0 m3. Under these conditions,water does not behave as an ideal gas.
System now undergoes a reverisble process in which it iscompressed to 100bar. The pressure-volume relationship is given by:Pv1.5=constant What is the final temperature andinternal energy of the system?Sketch process on PV diagram. Draw area thatrepresents the work of the process. Calculate the work done duringthe process. How much heat was exchanged?

Explanation / Answer

1) (a) Generally work done ON the gas by the surrounding is given by the integral: W = - ?Vi?Vf p dV (Vi?Vf denotes initial to final state) For constant pressure process this simplifies to W = - p · ?Vi?Vf dV = - p · (Vf - Vi) = p · (Vi - Vf) Having in mind that Pa·m³ = J it convert to these Si units before calculation: => W = 101325Pa · (12.0m³ - 16.2m³) = -425565J = -425.6kJ Work done by the gas to the surrounding has is the same but has the opposite sign. So just switch the sign and you're done. (b) Heat transferred to the gas is Q = 1000kcal = 4184kJ Internal energy change equals work done On the gas plus heat Transferred to the gas ?U = Q + W = 4184kJ - 425.6kJ = 3758.4kJ (2) (a) W = - ?Vi?Vf p dV = 0 since the volume does not change (b) ?U = Q + W = 269kJ + 0kJ = 269kJ (3) (a) adiabatic process means no heat is exchanged. So Q = 0 (b) ?U = Q + W = 0kJ + 1340kJ = 1340kJ (c) rise Internal energy of an ideal gas is solely a function of temperature: U = n·Cv·T (n number of moles Cv heat capacity at constant volume)

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