A cylinder that has a 40.0-cm radius and is 50.0 cm deep is filled with air at 2

Question

A cylinder that has a 40.0-cm radius and is 50.0 cm deep is filled with air at 20.0 degree C and 1.00 aim (Fig. P16.74a). A 20.0-kg piston is now lowered into the cylinder, compressing the air trapped inside as it takes equilibrium height h_i (Fig. P16.74b). Finally, a 25.0-kg dog stands on the piston, further compressing the air, which remains at 20 degree C (Fig. P16.74c). (a) How far down (Delta h) does the piston move when the dog steps onto It? (b) To what temperature should the gas be waned to raise the piston and dog back to h_i?

Explanation / Answer

Area of cross section(A)= 3.14 *(0.4)2

when there was only piston.pressure (P2)=101325 + 20*9.8/A=101714.93 Pa

P1= 101325 Pa,V1= A H1,V2=AH2

From P1V1=P2V2 we get H2= 49.81 cm

Similarly when there was Piston + Dog (P3)=101325 + (20+25)*9.8/A=102202.34 Pa

FRom Boyle' s Law again we get P1V1=P3V3 , V3=AH3 ,So H3=49.57 cm

(a) the piston goes (49.81-49.57) cm=0.24 cm inside after the dog steps

(b)From V1/T1=V3/T3  V1= A* 50 ,V3=A* 49.57,T1= ? ,T3=293 K

So T1=295.54 K=22.54 degree celcius

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