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CALC During the time 0.305 mol of an ideal gas under- goes an isothermal compression at \(22.0^{\circ} \mathrm{C}, 468 \mathrm{J}\) of work is done on it by the surroundings. (a) If the final pressure is 1.76 atm, what was the initial pressure? (b) Sketch a pV-diagram for the process.

Short Answer

Expert verified
(a) Initial pressure was approximately 0.950 atm. (b) The pV diagram is a hyperbolic curve showing decreasing volume and increasing pressure.

Step by step solution

01

Understand the Process

In an isothermal process, the temperature remains constant. From the given problem, we know that the ideal gas undergoes isothermal compression at 22.0°C, 468 J of work is done on it, and the final pressure is 1.76 atm.
02

Express Work Done in Terms of Pressures

For an isothermal compression of an ideal gas, the work done is given by the formula: \( W = nRT \ln\left(\frac{P_f}{P_i}\right) \), where \( n \) is the number of moles, \( R \) is the ideal gas constant (8.314 J/mol·K), \( T \) is the temperature in Kelvin, \( P_f \) is the final pressure, and \( P_i \) is the initial pressure.
03

Convert Temperature to Kelvin

Convert the temperature from Celsius to Kelvin: \( T = 22.0 + 273.15 = 295.15 \text{ K} \).
04

Substitute Values into the Work Formula

Given \( n = 0.305 \) mol, \( W = 468 \) J, \( P_f = 1.76 \) atm (note: 760 torr = 1 atm), substitute the values into the formula: \[ 468 = 0.305 \times 8.314 \times 295.15 \times \ln\left(\frac{1.76}{P_i}\right) \]
05

Solve for Initial Pressure \( P_i \)

First solve for \( \ln\left(\frac{1.76}{P_i}\right) \): \[ \ln\left(\frac{1.76}{P_i}\right) = \frac{468}{0.305 \times 8.314 \times 295.15} \]Calculate this expression to find:\[ \ln\left(\frac{1.76}{P_i}\right) = 0.617 \]Now rearrange to solve for \( P_i \):\[ \frac{1.76}{P_i} = e^{0.617} \]\[ P_i = \frac{1.76}{e^{0.617}} \]Finally, calculate \( P_i \) in atm.
06

Calculate\( P_i \)

Calculating the right-hand side, we have:\[ P_i = \frac{1.76}{1.853} \approx 0.950 \text{ atm} \]
07

Sketch the pV Diagram

Make a graph with volume \( (V) \) on the x-axis and pressure \( (P) \) on the y-axis. An isothermal compression will show a hyperbolic curve from a lower pressure \( (P_i = 0.950 \text{ atm}) \) to a higher pressure \( (P_f = 1.76 \text{ atm}) \) with decreasing volume.

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Key Concepts

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

Isothermal Process
An isothermal process is a fascinating concept where the temperature of the system remains constant throughout the process. In the world of thermodynamics, such processes are vital because they simplify calculations by maintaining a stable state.
To imagine an isothermal process, picture a gas in a sealed container being slowly compressed or expanded without changing its temperature. The heat transfer with the surroundings allows the gas to adjust, so the temperature stays the same. This means all the energy changes happen due to pressure and volume alterations, not temperature shifts.
  • For ideal gases, where the Ideal Gas Law applies, the relationship \( PV = nRT \) helps us understand how pressure and volume can change while temperature stays constant.
  • Since the temperature doesn't change, the internal energy of an ideal gas under an isothermal process remains unchanged as well.
In practical applications, isothermal processes are often used in engines and refrigerators, where maintaining consistent temperature is key for efficient operation.
Pressure Conversion
Pressure conversion is essential in solving many thermodynamics problems since the pressures can be given in different units. In our example, the pressures were given in atmospheres (atm), but converting between different units could be crucial depending on the region or context.

Common units of pressure include:
  • Pascal (Pa)
  • Atmosphere (atm), where 1 atm equals 101,325 Pa
  • Bar, where 1 bar equals 100,000 Pa
  • Millimeter of Mercury (mmHg) or Torr, where 1 atm equals 760 mmHg
Knowing how to convert between these different units allows for more versatile problem-solving approaches.
Let's say you find a pressure in mmHg but need it in atm for a calculation using the ideal gas law. You would simply use the conversion factor: \( 1 ext{ atm} = 760 ext{ mmHg} \). By this, you can adjust the pressure values seamlessly between the units.
Work-Energy Principle
The work-energy principle is a key idea in physics that ties together concepts of work done and energy change. When dealing with gases, especially during processes like compression or expansion, this principle provides a framework to understand energy transformations.

For an ideal gas undergoing an isothermal compression or expansion, we calculate the work done using the formula: \[ W = nRT \ln\left(\frac{P_f}{P_i}\right) \] Where:
  • \( W \) is the work done on the gas
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant
  • \( T \) is absolute temperature in Kelvin
  • \( P_f \) and \( P_i \) are the final and initial pressures respectively
This formula derives from integrating the pressure-volume work relationship while ensuring that the temperature remains constant.
An interesting aspect of the work-energy principle here is how it connects thermodynamics with mechanical systems. The energy input or output from system interactions causes changes in mechanical work, which is a measurable form of energy transfer.

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Most popular questions from this chapter

Starting with 2.50 mol of \(\mathrm{N}_{2}\) gas (assumed to be ideal) in a cylinder at 1.00 atm and \(20.0^{\circ} \mathrm{C},\) a chemist first heats the gas at constant volume, adding \(1.52 \times 10^{4} \mathrm{J}\) of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. (a) Calculate the final temperature of the gas. (b) Calculate the amount of work done by the gas. (c) Calculate the amount of heat added to the gas while it was expanding. (d) Calculate the change in internal energy of the gas for the whole process.

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 cylinder contains 0.250 mol of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\) . The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 Atm on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\) . Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) -diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

An air pump has a cylinder 0.250 \(\mathrm{m}\) long with a mov- able piston. The pump is used to compress air from the atmosphere (at absolute pressure \(1.01 \times 10^{5}\) Pa) into a very large tank at \(4.20 \times 10^{5}\) Pa gauge pressure. (For air, \(C_{V}=20.8 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) ) (a) The piston begins the compression stroke at the open end of the cylinder. How far down the length of the cylinder has the piston moved when air first begins to flow from the cylinder into the tank? Assume that the compression is adiabatic. (b) If the air is taken into the pump at \(27.0^{\circ} \mathrm{C}\) , what is the temperature of the compressed air? (c) How much work does the pump do in putting 20.0 mol of air into the tank?

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.

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