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A 25 kg piston is above a gas in a long vertical cylinder. Now the piston is released from rest and accelerates up in the cylinder reaching the end 5 m higher at a velocity of 25 m/s. The gas pressure drops during the process, so the average is 600 kPa with an outside atmosphere at 100 kPa. Neglect the change in gas kinetic and potential energy, and find the needed change in the gas volume.

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Answer:

[tex]\Delta V_{gas}=0.018m^3[/tex]

Explanation:

We define as,

c.V Piston

[tex](E_2-E_1)_{Pist}=m(u_2-u_1)+m[\frac{1}{2}V^2_2-0]+mg(h_2-0)[/tex]

[tex](E_2-E_1)_{Pist}=0+25*0.5*25^2+25*9.8*5[/tex]

[tex](E_2-E_1)_{Pist}=7712+5+1225.8=9038.3J[/tex]

The energy equation for the piston is

[tex]E_2-E_1=W_{gas}-W_{atm}=P_{avg}\Delta V_{gas}-P_0 \Delta V_{gas}[/tex]

Remember that

[tex]\Delta V_{atm}=-\Delta V_{gas}[/tex]

So,

[tex]\Delta V_{gas}=9.038kJ/(600-100)=0.018m^3[/tex]

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