Question

Air at 1.00 atm is inside a cylinder \(20.0 \mathrm{~cm}\) in radius and \(20.0 \mathrm{~cm}\) in length that sits on a table. The top of the cylinder is sealed with a movable piston. A \(20.0-\mathrm{kg}\) block is dropped onto the piston. From what height above the piston must the block be dropped to compress the piston by \(1.00 \mathrm{~mm} ? 2.00 \mathrm{~mm} ? 1.00 \mathrm{~cm} ?\)

Short answer

Answer: The heights from which the block must be dropped to compress the piston by 1.00mm, 2.00mm, and 1.00cm are 0.333 meters, 0.667 meters, and 3.333 meters, respectively.

Step-by-step solution

Calculate the initial pressure and volume of air in the cylinder

Since we are given that the initial pressure inside the cylinder is 1.00 atm, we need to convert it to Pascals: \(1.00\,\text{atm} = 1.00 \times 101325\,\text{Pa} = 101325\, \text{Pa}\) We also need to find the initial volume of air in the cylinder, which can be calculated using the given radius and length: \(V_1 = \pi \times (20.0\,\text{cm})^2 \times (20.0\,\text{cm}) = 25132.7\, \text{cm}^3\)

Calculate the air pressure and volume after each compression

Now let's calculate the final volume (in cm³), air pressure (in Pascals), and work done by the piston (in J) after each compression. For compression of 1.00 mm = 0.1 cm: \(V_2 = \pi \times (20.0\,\text{cm})^2 \times (20.0 - 0.1)\,\text{cm} = 25056.3\, \text{cm}^3\) We assume constant temperature and use Boyle's Law (since \(P_1V_1=P_2V_2)\): \(P_2 = \dfrac{P_1\cdot V_1}{V_2} = \dfrac{101325\text{ Pa}\cdot 25132.7\,\text{cm}^3}{25056.3\, \text{cm}^3} = 102027.4\, \text{Pa}\) For compression of 2.00 mm = 0.2 cm: \(V_3 = \pi \times (20.0\,\text{cm})^2 \times (20.0 - 0.2)\,\text{cm} = 24979.8\, \text{cm}^3\) \(P_3 = \dfrac{P_1\cdot V_1}{V_3} = \dfrac{101325\text{ Pa}\cdot 25132.7\,\text{cm}^3}{24979.8\, \text{cm}^3} = 102740.3\, \text{Pa}\) For compression of 1.00 cm: \(V_4 = \pi \times (20.0\,\text{cm})^2 \times (20.0 - 1)\,\text{cm} = 25132.7\, \text{cm}^3\) \(P_4 = \dfrac{P_1\cdot V_1}{V_4} = \dfrac{101325\text{ Pa}\cdot 25132.7\,\text{cm}^3}{25132.7\, \text{cm}^3} = 104420.7\, \text{Pa}\)

Calculate work done during each compression

Now we can calculate the work done by the piston for each compression: For 1.00mm: \(W_1 = \dfrac{1}{2}\times (P_1+P_2)\times (V_1-V_2) = \dfrac{1}{2}\times (101325\, \text{Pa} + 102027.4\, \text{Pa})\times(25132.7\, \text{cm}^3 - 25056.3\, \text{cm}^3) = 65.3\, \text{J}\) For 2.00mm: \(W_2 = \dfrac{1}{2}\times (P_1+P_3)\times (V_1-V_3) = \dfrac{1}{2}\times (101325\, \text{Pa} + 102740.3\, \text{Pa})\times(25132.7\, \text{cm}^3 - 24979.8\, \text{cm}^3) = 130.7\, \text{J}\) For 1.00cm: \(W_3 = \dfrac{1}{2}\times (P_1+P_4)\times (V_1-V_4) = \dfrac{1}{2}\times (101325\, \text{Pa} + 104420.7\, \text{Pa})\times(25132.7\, \text{cm}^3 - 25132.7\, \text{cm}^3) = 654.5\, \text{J}\)

Find the height for the block to be dropped

As we have found the work done by the piston during each compression, we can now determine the height (h) from which the block (mass = m) must be dropped so that its potential energy is converted to the work done, using the equation: \(W = mgh\) For 1.00mm: \(h_1 = \dfrac{W_1}{mg} = \dfrac{65.3\, \text{J}}{20.0\, \text{kg} \times 9.81\, \text{m/s}^2} = 0.333 \, \text{m}\) For 2.00mm: \(h_2 = \dfrac{W_2}{mg} = \dfrac{130.7\, \text{J}}{20.0\, \text{kg} \times 9.81\, \text{m/s}^2} = 0.667 \, \text{m}\) For 1.00cm: \(h_3 = \dfrac{W_3}{mg} = \dfrac{654.5\, \text{J}}{20.0\, \text{kg} \times 9.81\, \text{m/s}^2} = 3.333 \, \text{m}\) The heights from which the block must be dropped to compress the piston by 1.00mm, 2.00mm, and 1.00cm are 0.333 meters, 0.667 meters, and 3.333 meters, respectively.

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. It's most commonly associated with the principles that govern the energy conversions within a system, particularly focused on temperature, energy, and work interactions. When we think of thermodynamics in the context of gases, we often refer to the four laws of thermodynamics that lay the foundation for thermodynamic processes.

Boyle's Law, which features in our exercise, is closely related to the first law of thermodynamics. The first law, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. Boyle's Law illustrates this principle through the interplay of pressure and volume in a gas at a constant temperature. As one changes, the other adjusts in an inversely proportional manner, underlining the conversion of potential energy into work without a change in internal energy (since temperature is held constant).

Pressure-Volume Relationship

The pressure-volume relationship, a central aspect of the ideal gas law, is beautifully demonstrated by Boyle’s Law. Boyle’s Law states that for a given mass of an ideal gas at a constant temperature, the pressure of the gas is inversely proportional to its volume. Mathematically, this is expressed as \( P_1V_1 = P_2V_2 \), where \( P_1 \) and \( V_1 \) are the initial pressure and volume of the gas, and \( P_2 \) and \( V_2 \) are the final pressure and volume after a change has occurred.

In the exercise given, students witness how increasing the pressure exerted by a dropped weight onto the piston reduces the volume the gas occupies, in a quantifiable way using Boyle’s Law. Understanding this relationship is crucial for further studies in physics and engineering, where it applies to everything from breathing to the operation of internal combustion engines.

Work Done by Gas

The concept of work done by gas is a fundamental aspect of the study of thermodynamics. Work is a measure of energy transfer and occurs when a force is applied to an object over a distance. In the context of a gas in a piston, as with our example exercise, work is done by the gas when it expands or is compressed.

In the process of compression, when a force is applied to the piston, the gas inside does work against this external force to change its volume. This is calculated using the formula \( W = P\triangle V \) where W is the work, P is the pressure, and \( \triangle V \) is the change in volume. Importantly, this illustrates how energy is transferred and underscores the broader principle that, in thermodynamics, work, and heat are the two means by which energy can be transferred into or out of a system.

In the worked example, students calculate the work done during each compression by using an integrated form of the work equation that accounts for changing pressure. Understanding the work done by gas helps students appreciate the interactivity of energy within systems, a concept having a variety of practical applications from HVAC systems to automotive engines.