91Ó°ÊÓ

A piston-fitted cylinder with a 6 -cm inner diameter contains \(1.40 \mathrm{g}\) of nitrogen. The mass of the piston is 4.50 kg, and a 25.00-kg weight rests on the piston. The gas temperature is 30^ C, and the pressure outside the cylinder is 2.50 atm. (a) Prove that the absolute pressure of the gas in the cylinder is \(3.55 \times 10^{5} \mathrm{Pa}\). Then calculate the volume occupied by the gas, assuming ideal- gas behavior. (b) Suppose the weight is abruptly lifted and the piston rises to a new equilibrium position. Further suppose that the process takes place in two steps: a rapid step in which a negligible amount of heat is exchanged with the surroundings, followed by a slow step in which the gas returns to \(30^{\circ} \mathrm{C}\). Considering the gas as the system, write the energy balances for step \(1,\) step \(2,\) and the overall process. In all cases, neglect \(\Delta E_{\mathrm{k}}\) and \(\Delta E_{\mathrm{p}} .\) If \(\tilde{U}\) varies proportionally with \(T\), does the gas temperature increase or decrease in step 1? Briefly explain your answer. (c) The work done by the gas equals the restraining force (the weight of the piston plus the force due to atmospheric pressure) times the distance traveled by the piston. Calculate this quantity and use it to determine the heat transferred to or from (state which) the surroundings during the process.

Step by step solution

01

Calculation of the Absolute Pressure

The given mass of nitrogen is 1.4g, which we'll convert to moles i.e., \(n = \frac{1.4g}{28.0134g/mol} = 0.05mol\). The cylinder has a radius of 3 cm, which we convert to meters, so \(r = 0.03m\). The total force exerted is due to the weight placed on the piston, the piston itself, and the atmospheric pressure. Thus, \(F = (4.5kg + 25kg)9.8 m/s^2 + 2.5atm * 1.013 x 10^5 Pa/m^2 * \pi r^2\). Since Pressure = Force/Area, \(P = \frac{F}{\pi r^2}\).

02

Calculation of the Volume of Gas

Volume of the gas can be calculated using the ideal gas law. \(PV = nRT\), where P is the absolute pressure calculated from the previous step, n = number of moles, R = 8.314 J/mol.K (Universal Gas Constant) and T is the temperature in Kelvin. Convert the given temperature of 30°C to Kelvin by adding 273.15. The volume \(V\) can now be calculated.

03

Energy Balance for the rapid step and slow step

Ignoring changes in kinetic and potential energy, the first law of thermodynamics in terms of internal energy is \(\Delta U = Q - W\), where Q is the heat transferred and W is work done on the system by the surroundings. In the rapid step, heat exchange is negligible, hence Q (heat transferred to the system) = 0, and \(\Delta U = -W\). In the slow step, the system returns to its initial temperature, hence \(\Delta U = 0\) and \(Q = W\). As a result, the system's temperature increases in the first step (as energy is added to the system as work and goes into increasing the internal energy and thereby the temperature), then decreases back to the initial value in the second step.

04

Work Done and Heat Transferred

The work done by the gas equals the restraining force times the distance traveled by the piston. We can write an equation for this. From the first law of thermodynamics, heat transferred can also be determined. From the work done by the system, the sign of the heat transfer can be deduced. A positive work value indicates that the process is an expansion (doing work on the surroundings) and resultant heat will be released (exothermic), while a negative work value indicates a compression (work is done on the system) and the resultant reaction will absorb heat (endothermic).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Balance

In thermodynamics, energy balance is a crucial principle that helps us understand how energy flows in and out of a system. The principle states that the change in internal energy (\( \Delta U \)) of a system is equal to the heat added to the system (\( Q \)) minus the work done by the system (\( W \)). This is mathematically expressed as:
\[ \Delta U = Q - W \]
In the presented problem, the concept of energy balance is applied during two steps when the gas expands rapidly and then returns to equilibrium slowly. During the rapid step, heat exchange is negligible, which simplifies the energy balance to:
\[ \Delta U = -W \]
In the slow step, since the gas returns to its initial temperature, we have an energy balance of:
\[ Q = W \]
This understanding is key in determining how internal energy and consequently temperature changes.

Thermodynamics

Thermodynamics is the science of energy and its various manifestations, including heat, work, and internal energy. It provides a framework to analyze how energy scales and interplays within a system. Its principles are used to predict and explain the behavior of gas in this exercise.
For the exercise at hand, the first law of thermodynamics is pivotal. This law links changes in internal energy to heat exchange and work done. It is elegantly captured in the formula:
\[ \Delta U = Q - W \]
Thermodynamic processes can be categorized by how they transfer heat and work. Our problem involves an adiabatic-like rapid step with no heat transfer (step 1), followed by a step where the gas temperature normalizes, at the end exerting work equal to the heat transferred (step 2).
Using these laws helps students predict how systems like the piston and gas will behave when external conditions change. Understanding these concepts provides a window into the workings of combustion engines, refrigeration, and more.

Internal Energy

Internal energy is the total energy contained within a system due to the random motions and interactions of its molecules. It is a function of the state variables of the system such as temperature and pressure.
In the context of the exercise, when the weight on the piston is lifted, the internal energy changes during the rapid expansion step. As the system absorbs work energy:

• This results in an increase in internal energy.
• This increase is because the work energy is stored as internal energy in the expanded system, causing the temperature to rise momentarily.

In the second step, where the system is allowed to reach thermodynamic equilibrium at its initial temperature, the internal energy returns to its original value. This makes internal energy a 'state function', dependent only on the state and not how it got there.
Understanding internal energy is fundamental in thermodynamics, aiding the comprehension of how systems absorb or release energy and the effects on temperature and state.