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Chapter 19: Problem 1

Two moles of an ideal gas are heated at constant pressure from \(T=27^{\circ} \mathrm{C}\) to \(T=107^{\circ} \mathrm{C}\) (a) Draw a p \(V\) -diagram for this process. (b) Calculate the work done by the gas.

Short Answer

Expert verified
Work done by the gas is 1328.64 J.

Step by step solution

01

Understanding the Process

We have two moles of an ideal gas undergoing a process at constant pressure where the temperature increases from \(27^{\circ} C\) to \(107^{\circ} C\). At constant pressure, the volume will increase as the temperature increases.
02

Convert Temperatures to Kelvin

To use the ideal gas law, we need to convert the temperatures from Celsius to Kelvin. \(T_1 = 27^{\circ} C + 273 = 300\, K\) and \(T_2 = 107^{\circ} C + 273 = 380\, K\).
03

Draw the p-V Diagram

Since the pressure is constant, the process is represented as a horizontal line in the p-V diagram. The initial point is at \(T_1=300\, K\) and the final point is at \(T_2=380\, K\). The line moves to the right, indicating an increase in volume.
04

Calculate the Work Done by the Gas

The work done by the gas at constant pressure is calculated using the formula: \[ W = P(V_2 - V_1) \]Using the ideal gas law, \( PV = nRT \), we find the volumes:\[ V_1 = \frac{nRT_1}{P} \] and \[ V_2 = \frac{nRT_2}{P} \]Substituting these into the work formula gives:\[ W = P\left(\frac{nRT_2}{P} - \frac{nRT_1}{P}\right) = nR(T_2 - T_1) \]With \(n = 2\, \text{moles}\) and \(R = 8.314\, \text{J/mol鈰匥}\), compute:\[ W = 2 \cdot 8.314 \cdot (380 - 300) = 2 \cdot 8.314 \cdot 80 \]\[ W = 1328.64 \, \text{J} \]
05

Verify the Units and Results

Ensure all steps followed, confirm the calculations of temperature conversion, work done, and appropriate units used in equations.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental principle in thermodynamics. It combines several separate gas laws into one equation: \(PV = nRT\). Here, \(P\) stands for pressure, \(V\) for volume, \(n\) for the number of moles of gas, \(R\) for the ideal gas constant (8.314 J/mol鈰匥), and \(T\) for temperature in Kelvin.

This law is used to relate the properties of an ideal gas, which is a theoretical gas composed of many randomly moving, non-interacting point particles. Real gases approximate ideal behavior under many conditions, making the ideal gas law very useful in practical applications.

To effectively use this law, remember to always convert temperature to Kelvin and use consistent units throughout the calculations.
Work Done by Gas
In thermodynamics, the concept of work done by gas is crucial in understanding energy transfers. When gas expands, or a system does work on the surroundings, the energy transfer involves work.

Work done by gas at constant pressure can be calculated using the formula: \[ W = P(V_2 - V_1) \]
Here, \(W\) is the work done, \(P\) is the constant pressure, and \(V_2 - V_1\) is the change in volume.
  • If the gas expands, work is done by the gas.
  • If the gas is compressed, work is done on the gas.
The work done by gas is generally measured in Joules. Proper understanding of this concept helps in applications such as engines, refrigerators, and atmospheric studies.
PV Diagram
A PV diagram, or Pressure-Volume diagram, graphically represents the process a gas undergoes during transformation. It plots pressure (P) on the y-axis and volume (V) on the x-axis.

In the context of a constant pressure process, the diagram displays as a horizontal line. The process ensures the pressure remains the same while the volume may change with variations in temperature.

Understanding how to read a PV diagram is vital for interpreting the mechanical work done during processes involving gases. You can visually identify processes such as isobaric (constant pressure), isochoric (constant volume), and isothermal (constant temperature) transformations.
Constant Pressure Process
A constant pressure process, also known as an isobaric process, occurs when a gas undergoes changes with pressure remaining unchanged. This property simplifies many calculations in thermodynamics.
  • Volume increases with a rise in temperature.
  • Such processes are useful for understanding real-world applications like heating and cooling systems.
In our specific scenario, when the temperature increases from \(27^{\circ} C\) to \(107^{\circ} C\), the volume also increases, allowing the gas to perform work.

Constant pressure conditions make it directly feasible to calculate work done by using the formula \(W = nR(T_2 - T_1)\), integrating the effects of a changing temperature on volume.

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Most popular questions from this chapter

In an adiabatic process for an ideal gas, the pressure decreases. In this process does the internal energy of the gas increase or decrease? Explain your reasoning.

During an isothermal compression of an ideal gas, 335 \(\mathrm{J}\) of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

A monatomic ideal gas expands slowly to twice its original volume, doing 300 \(\mathrm{J}\) of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is (a) isothermal; (b) adiabatic; (c) isobaric.

A Thermodynamic Process in an Insect. The African bombardier beetle (Stenaptinus insignis) can emit a jet of defensive spray from the movable tip of its abdomen (Fig. Pl9.57). The beetle's body has reservoirs of two different chemicals; when the beetle is disturbed, these chemicals are combined in a reaction chamber, producing a compound that is warmed from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 \(\mathrm{m} / \mathrm{s}(68 \mathrm{km} / \mathrm{h})\) , scaring away predators of all kinds. (The beetle shown in the figure is 2 \(\mathrm{cm}\) long.) Calculate the heat of reaction of the two chemicals (in \(\mathrm{J} / \mathrm{kg} ) .\) Assume that the specific heat of the two chemicals and the spray is the same as that of water, \(4.19 \times 10^{3} \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and that the initial temperature of the chemicals is \(20^{\circ} \mathrm{C}\) .

Five moles of an ideal monatomic gas with an initial temperature of \(127^{\circ} \mathrm{C}\) expand and, in the process, absorb 1200 \(\mathrm{J}\) of heat and do 2100 \(\mathrm{J}\) of work. What is the final temperature of the gas?
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