Air at 101 kPa is trapped in a piston-cylinder deviceoriented horizontally. Init

Question

Air at 101 kPa is trapped in a piston-cylinder deviceoriented horizontally. Initially the volume in the cylinderis 2.00 liters and the face of the piston is at x=0, a position atwhich the spring attached to the piston exerts noforce.    Atmospheric pressure is 101 kPa and thearea of the piston face is 0.018 m2.   The airexpands slowly until its volume is 3.00 liters.   Thespring is linear (F =kx) with a spring constant k = 16.2kN/m.   Determine the final pressure of the air and thework done by the air on the piston.

Explanation / Answer

A= 0.018 m^2 P1= 101 kPa Po = 101 kPa V1= 2 L = 0.002 m^3 V2= 3 L = 0.003 m^3 k= 16.2 *10^3 (N/m) P2 = ??     P2*A = Fs + Po*A      V = V2-V1 = A*x    0.003 - .002 = A*x       x= 0.0555 m    Fs = k*x = 16200 * 0.0555 = 899.1 N     P2*A = Fs + Po*A     P2*0.018 = 899.1 + 101*10^3 * 0.018 (P2 = 151 kPa) W = Wb + Ws W = (0.003-.002)*(101*10^3) +0.5*(0.003-0.002)*(151*10^3-101*10^3) (W = 0.126 KW) Good luck

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