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When a gas undergoes a quasi-static isobaric change in volume from 10.0 to \(2.0 \mathrm{L}, 15 \mathrm{J}\) of work from an external source are required. What is the pressure of the gas?

Short Answer

Expert verified
The pressure of the gas during this quasi-static isobaric process is approximately \(1875 \, Pa\).

Step by step solution

01

Write down the given information

Initial volume V1: \(10.0 \, L\) Final volume V2: \(2.0 \, L\) Work done W: \(15 \, J\)
02

Calculate the change in volume

The change in volume is calculated as: \[\Delta V = V2 - V1\] Substitute the given values: \[\Delta V = 2.0 \, L - 10.0 \, L = -8.0 \, L\]
03

Rearrange the formula for pressure

The formula for work done in a quasi-static isobaric process is: \[W = P \cdot \Delta V\] We want to find the pressure, so rearrange the formula to solve for \(P\): \[P = \frac{W}{\Delta V}\]
04

Plug in the given values and calculate the pressure

Now, substitute the given values for work done and change in volume into the formula: \[P = \frac{15 \, J}{-8.0 \, L}\] Divide the numbers: \[P \approx -1.875 \, \frac{J}{L}\] Since pressure is a scalar quantity, we can ignore the negative sign and state the pressure in its appropriate units. Convert the units to \(Pa (Pascals)\) as: \[1 \, \frac{J}{L} = 1 \, \frac{Nm}{m^3} = 1000 \, \frac{N}{m^3} = 1000 \, Pa\] Finally, multiply the pressure by the conversion factor: \[P \approx 1.875 \times 1000 \, Pa = 1875 \, Pa\] The pressure of the gas during this quasi-static isobaric process is approximately \(1875 \, Pa\).

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Key Concepts

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

Quasi-Static Process
Understanding thermodynamic processes is crucial for grasping the principles of how energy is transferred in systems such as gases. A quasi-static process is one that is carried out sufficiently slowly to allow the system to remain in internal equilibrium. In simpler terms, we can think of it as happening in such slow motion that, at every moment, the system's properties (pressure, volume, temperature) are nearly uniform throughout.

This characteristic allows us to apply the laws of thermodynamics very accurately, as it eliminates the gradients that could cause non-equilibrium conditions. Consider inflating a balloon very slowly; a quasi-static process would mean that the pressure inside and outside the balloon remains almost the same throughout the process, avoiding rapid changes and stresses.
Work Done by Gas
When we talk about work done by a gas or on a gas, we're referring to the energy transferred when a gas changes volume under pressure. During an isobaric process, where the pressure is constant, the work done can be represented by a simple equation:

\[ W = P \times \triangle V \]
Here, \( W \) is the work done, \( P \) is the constant pressure, and \( \triangle V \) is the change in volume. The sign of the work tells us who is doing the work: if it's positive, work is done on the gas (like compressing it), and if it's negative, the gas is doing work (like expanding against an external pressure). In educational settings, real-life examples such as a piston moving inside a cylinder help visualize this concept.
Pressure Calculation
Pressure is a fundamental concept in physical sciences, defined as force per unit area. It tells us how much force is exerted over a specific area and plays a vital role in understanding gas behaviors. Calculating the pressure of a gas in a quasi-static isobaric process can be achieved by rearranging the work formula mentioned earlier:

\[ P = \frac{W}{\triangle V} \]

What Determines Pressure?

  • Volume Change: The degree of volume change affects the pressure calculation. When the volume decreases, the pressure usually increases (if the amount of gas and temperature remains constant).
  • Work Done: This relates to the energy required to bring about the volume change. The more work done on the gas, the more pressure is applied to it.

It's important to note that the units you use for work done and volume change must be coherent to get a correct value for pressure.
Gas Volume Change
Gas volume change, especially in an isobaric process, is an indicator of energy transfer through work. Here's what students should remember about volume changes in gases:
  • In an isobaric process, if a gas expands, it does work on the surroundings, leading to a negative work value.
  • If a gas is compressed, the surroundings do work on it, and this work value is positive.
  • The magnitude of the volume change is crucial for calculating the work done and, subsequently, the pressure during the process, as seen in the textbook example.

To improve student's understanding, visual aids like graphs showing volume versus pressure for isobaric processes can provide a clearer picture of how gas volume changes under constant pressure.

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Most popular questions from this chapter

An ideal gas has a pressure of 0.50 atm and a volume of 10 L. It is compressed adiabatically and quasi-statically until its pressure is 3.0 atm and its volume is 2.8 L. Is the gas monatomic, diatomic, or polyatomic?

There is no change in the internal energy of an ideal gas undergoing an isothermal process since the internal energy depends only on the temperature. Is it therefore correct to say that an isothermal process is the same as an adiabatic process for an ideal gas? Explain your answer.

An ideal gas expands isothermally along AB and does 700 J of work (see below). (a) How much heat does the gas exchange along AB? (b) The gas then expands adiabatically along BC and does 400 J of work. When the gas returns to A along CA, it exhausts 100 J of heat to its surroundings. How much work is done on the gas along this path?

The van der Waals coefficients for oxygen are \(a=0.138 \mathrm{J} \cdot \mathrm{m}^{3} / \mathrm{mol}^{2}\) and \(b=3.18 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol}\). Use these values to draw a van der Waals isotherm of oxygen at \(100 \mathrm{K}\). On the same graph, draw isotherms of one mole of an ideal gas.

One mole of an ideal gas is initially in a chamber of volume \(1.0 \times 10^{-2} \mathrm{m}^{3}\) and at a temperature of \(27^{\circ} \mathrm{C}\) (a) How much heat is absorbed by the gas when it slowly expands isothermally to twice its initial volume? (b) Suppose the gas is slowly transformed to the same final state by first decreasing the pressure at constant volume and then expanding it isobarically. What is the heat transferred for this case? (c) Calculate the heat transferred when the gas is transformed quasi-statically to the same final state by expanding it isobarically, then decreasing its pressure at constant volume.

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