Question

A vertical 12 -cm diameter piston-cylinder device contains an ideal gas at the ambient conditons of 1 bar and \(24^{\circ} \mathrm{C} .\) Initially, the inner face of the piston is \(20 \mathrm{cm}\) from the base of the cylinder. Now an external shaft connected to the piston exerts a force corresponding to a boundary work input of \(0.1 \mathrm{kJ} .\) The temperature of the gas remains constant during the process. Determine ( \(a\) ) the amount of heat transfer, (b) the final pressure in the cylinder, and ( \(c\) ) the distance that the piston is displaced.

Short answer

The final pressure in the piston-cylinder device after the process is 1.38 bar.

Step-by-step solution

Find the initial volume and mass of the gas

First, we need to find the initial volume of the gas in the piston-cylinder device, which can be calculated using the given diameter and height. For a cylinder, volume (V) is given by the formula: V = πr²h Diameter (d) of the piston is 12 cm, so the radius (r) will be 6 cm or 0.06 m. The initial height (h) of the gas is 20 cm or 0.2 m. Therefore, the initial volume (V1) of the gas will be: V1 = π(0.06)²(0.2) = 0.00361 m³. Now, we can find the mass of the gas using the ideal gas law, which states that: PV = mRT The gas constant for air (R) is 287 J/(kg·K). Given pressure (P1) is 1 bar or 100,000 Pa, and the temperature (T) is \(24^\circ C\) or 297.15 K. Hence, mass (m) of the gas can be found as: m = (P1 * V1) / (R * T) = (100,000 * 0.00361) / (287 * 297.15) = 0.1308 kg.

Find the work done and the final volume of the gas

Next, we need to find the work (W) done within the cylinder by the external force. The given boundary work input is 0.1 kJ or 100 J. We know that work done is given by: W = -P * ΔV As the temperature remains constant, the gas goes through an isothermal process. Therefore, from the ideal gas law, PV = constant. So, we can write: P1 * V1 = P2 * V2 Now, we can find the final volume (V2) of the gas using the equation: V2 = V1 - (W / P1) = 0.00361 - (100 / 100,000) = 0.00261 m³.

Calculate the final pressure and distance of displacement

Using the derived equation from Step 2, we can find the final pressure (P2) in the cylinder: P2 = P1 * (V1 / V2) = 100,000 * (0.00361 / 0.00261) = 138,077 Pa or 1.38 bar. Now, we can find the displacement (Δh) of the piston, which can be found using the final volume (V2): Δh = (V2 / (πr²)) - h = (0.00261 / (π * (0.06)²)) - 0.2 = -0.112 m or 11.2 cm. The negative sign indicates that the piston is moving downward.

Find the amount of heat transfer

We can use the first law of thermodynamics for an isothermal process (ΔU = 0) to find the heat transfer (Q), which states that: Q = W + ΔU In an isothermal process, the internal energy change (ΔU) is zero. Therefore, the heat transfer will be equal to the work done: Q = W = 100 J Now we have the answers for the exercise: (a) Amount of heat transfer, Q = 100 J. (b) Final pressure in the cylinder, P2 = 1.38 bar. (c) The distance that the piston is displaced, Δh = 11.2 cm (downward).

Key concepts

Ideal Gas Law

Understanding the ideal gas law is fundamental to solving problems involving a piston-cylinder device containing an ideal gas. This law attributes a simple, yet powerful relationship that connects pressure (P), volume (V), temperature (T), and mass (m) of a gas, which is represented by the equation PV = mRT. Here,

• P stands for the absolute pressure of the gas,
• V is the volume occupied by the gas,
• m is the mass of the gas,
• R is the specific gas constant, which varies for different gases,
• T is the absolute temperature in Kelvin.

The ideal gas law is essential in calculating properties of gases under different conditions. For example, in the step-by-step solution, by knowing the initial temperature, pressure, and volume, we're able to deduce the mass of the gas with the equation. Similarly, knowing the relationship PV = mRT helps us determine how volume and pressure change in relation to each other, which is particularly useful in isothermal processes where the temperature remains constant.

Isothermal Process

An isothermal process is a transformation that occurs at a constant temperature. This means that no matter how the gas is compressed or expanded within the piston-cylinder device, the temperature of the gas does not change throughout the process. When a gas undergoes an isothermal expansion or compression, the ideal gas law implies that the pressure and volume will inversely change while maintaining the product PV constant.
During isothermal process, any heat transfer to the gas results in work done by the gas and vice versa; therefore, if work is done on the gas, like in our example, an equal amount of heat is transferred from the gas to ensure temperature remains constant. This interplay between heat and work is what keeps the temperature stable, defining the nature of isothermal processes.

Boundary Work

Boundary work is the energy transferred across the system boundary by an expanding or contracting volume under pressure. It is a crucial concept for understanding work interactions in thermodynamic systems, particularly with piston-cylinder setups. In the given solution, boundary work is performed by an external shaft connected to the piston, manifesting as the force the gas inside the cylinder uses to push the piston or the force exerted to compress the gas.
The math behind calculating this can be represented as W = -PΔV, which correlates to the work as the negative product of pressure and the change in volume. As demonstrated, knowing the boundary work can help deduce the final volume of gas after the process has taken place, and further, determine other important properties of the system.

Heat Transfer

Heat transfer is the movement of thermal energy due to temperature difference. In thermodynamics, when a system, like our gas in the piston-cylinder device, undergoes any type of change, heat transfer is one of the primary ways it can exchange energy with its surroundings.
In the context of the solution provided, the first law of thermodynamics, which is essentially energy conservation, is used to connect heat transfer with work done and the change in internal energy. As stated in the text, for an isothermal process, the internal energy change (ΔU) is zero because temperature remains constant. This leads us to the conclusion that the heat transfer (Q) equals the work done (W). Consequently, no heat is lost or gained by the gas in the cylinder and the entire work done on the system is used to transfer heat out of the system.