Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 16, Problem 74P

A cylinder that has a 40.0-cm radius and is 50.0 cm deep is filled with air at 20.0°C and 1.00 atm (Fig. P10.74a). A 20.0-kg piston is now lowered into the cylinder, compressing the air trapped inside as it takes equilibrium height hi (Fig. P16.74b). Finally, a 25.0-kg dog stands on the piston, further compressing the air, which remains at 20°C (Fig. P16.74c). (a) How far down (Δh) does the piston move when the dog steps onto it? (b) To what temperature should the gas be warmed to raise the piston and dog back to hi?

Chapter 16, Problem 74P, A cylinder that has a 40.0-cm radius and is 50.0 cm deep is filled with air at 20.0C and 1.00 atm

(a)

Expert Solution
Check Mark
To determine

The height which the piston moves down when the dog steps onto the piston.

Answer to Problem 74P

The height at which the piston move down when the dog steps onto the piston is 2.38mm_.

Explanation of Solution

Write the expression for Boyle’s law, for the initial and final state since the temperature is constant.

    PV=P0V0        (I)

Here, P is the pressure after the compression, P0 is the initial pressure, V is the volume after the compression, and V0 is the initial volume.

Use Ahi for V, and Ah0 for V0 in the equation (I), and solve for P.

    PAhi=P0Ah0P=P0(h0hi)        (II)

Here, hi is the height after the first compression, and h0 is the initial height.

When the piston moves down, then the external pressure on the piston is mpgA. That is the final pressure is P=P0+mpgA.

Equate the equation (II) with the equation P=P0+mpgA.

    P0+mpgA=P0(h0hi)P0(1+mpgP0A)=P0(h0hi)hi=h01+mpgP0A        (III)

Replace the mass mp with total mass mp+M in the equation (III), to find final height of the piston.

    h=h01+(mp+M)gP0A        (IV)

Here, M is the mass of the dog.

The piston moves downward when the dog steps on the piston. It can be expressed as,

    Δh=hih        (V)

Use equation (III) and (IV) in (V).

    Δh=h01+mpgP0Ah01+(mp+M)gP0A        (VI)

Conclusion:

Substitute, 50.0m for h0, 20.0kg for mp, 9.80m/s2 for g, 1.013×105Pa for P0, π(0.400m)2 for A, and 25.0kg for M in the equation (VI), to find Δh.

    Δh=50.0m1+(20.0kg)(9.80m/s2)(1.013×105Pa)π(0.400m)250.0m1+(20.0kg+25.0kg)(9.80m/s2)(1.013×105Pa)π(0.400m)2=2.38mm

Therefore, the height at which the piston move down when the dog steps onto the piston is 2.38mm_.

(b)

Expert Solution
Check Mark
To determine

The temperature at which the gas should be warmed to raise the piston and dog back to hi.

Answer to Problem 74P

The temperature at which the gas should be warmed to raise the piston and dog back to hi is 21.4°C_.

Explanation of Solution

Write the expression for Charles’s law, for the initial and final state since the pressure is constant.

    VT=VTi        (VII)

Here, T is the temperature at which the gas should be warmed to raise the piston and dog back to hi, Ti is the temperature after the first compression.

Use Ahi for V, and Ah for V in the equation (VII), and solve for T.

    AhiT=AhTiT=Ti(hih)        (VIII)

Use equation (III) and (IV) in the equation (VIII).

    T=Ti(h01+mpgP0Ah01+(mp+M)gP0A)=Ti(1+(mp+M)gP0A1+mpgP0A)        (IX)

Conclusion:

Substitute, 293K for Ti, 50.0m for h0, 20.0kg for mp, 9.80m/s2 for g, 1.013×105Pa for P0, π(0.400m)2 for A, and 25.0kg for M in the equation (IX), to find T.

    T=293K[1+(20.0kg+25.0kg)9.80m/s2(1.013×105Pa)π(0.400m)21+20.0kg×9.80m/s2(1.013×105Pa)π(0.400m)2]=294.4K=294.4K273=21.4°C

Therefore, The temperature at which the gas should be warmed to raise the piston and dog back to hi is 21.4°C_.

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Answer for A, and the answer has to be in mm
A cylinder that has a 41.5-cm radius and is 50.0 cm deep is filled with air at 29.0°C and 1.00 atm shown in figure (a). A 24.0-kg piston is now lowered into the cylinder, compressing the air trapped inside as it takes equilibrium height hi as shown in figure (b). Finally, a 27.0-kg dog stands on the piston, further compressing the air, which remains at 29.0°C as shown in figure (c). When the dog steps onto it, the piston moves 2.35mm. To what temperature (in Celcius) should the gas be warmed to raise the piston and the dog back to hi?  (31.74°C is incorrect) Do you somehow have to find the new pressure of the piston after the dog stepped on it to find the temperature needed to raise the piston back up to its original height?
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Chapter 16 Solutions

Principles of Physics: A Calculus-Based Text

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