A certain fluid at 10 barias contained in cylinder behind apiston ,the initial v

Question

A certain fluid at 10 barias contained in cylinder behind apiston ,the initial volume being 0.05 m3 calculate thework done by the fluid when it expands reversibly i)according to a law PV= constant to a finalvolume of 0.1 m3 ii)according to a law PV3=constant to a final volume of 0.06 m3 iii)according to a law P=(A/V3 )-(B/V) ,toa final volume of 0.1 m3 and a pressure of 1bar,where A and B are constants. sketch all processes on P-Vdiagram A certain fluid at 10 barias contained in cylinder behind apiston ,the initial volume being 0.05 m3 calculate thework done by the fluid when it expands reversibly i)according to a law PV= constant to a finalvolume of 0.1 m3 ii)according to a law PV3=constant to a final volume of 0.06 m3 iii)according to a law P=(A/V3 )-(B/V) ,toa final volume of 0.1 m3 and a pressure of 1bar,where A and B are constants. sketch all processes on P-Vdiagram iii)according to a law P=(A/V3 )-(B/V) ,toa final volume of 0.1 m3 and a pressure of 1bar,where A and B are constants. sketch all processes on P-Vdiagram

Explanation / Answer

i). PV=constant. =X we have. dA=P*dV=X*dV/V integrate this we have. A=X*ln(V2/V1). =10*10e5*0,05*ln(0,1/0,05)=3,466e4(J) ii). similar to above we have. PV^3=constant = X. so dA=P*dV=X*dV/V^3. intergrate this we have. A=X*(1/V1^2-1/V2^2)/2=10*10e5*0,05^3*(1/0,05^2-1/0,06^2)/2= A=7640(J). iii). this part actually is the compose on two above part we have. P=A/V^3-B/V. so dA=P*dV=A*dV/V^3-B*dV/V. integrate this you should have the answer. (i think at least ushould try to do its). the diagram is draw by math only

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