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During the time 0.305 mol of an ideal gas undergoes an isothermal compression at 22.0C, 392 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) The initial pressure was approximately 1.87 atm. (b) Sketch the curve as hyperbolic from high volume, low pressure to low volume, high pressure on the pV-diagram.

Step by step solution

01

Understanding Isothermal Compression

In an isothermal process, the temperature remains constant. For an ideal gas undergoing isothermal compression, the internal energy doesn't change, meaning the work done on the gas is equal to the heat transferred out. We'll use the ideal gas equation to find the initial pressure.
02

Using the Ideal Gas Equation

For an ideal gas, the equation is given by PV=nRT. Since the temperature is constant, P1V1=P2V2 applies. The work done on the gas W=nRTln(V2V1) helps find the ratio V2V1. Use W=392 J and the gas constant R=8.314 J/mol K to calculate T in Kelvin, n=0.305 mol, and T=22.0+273.15 K=295.15 K.
03

Calculating Temperature in Kelvin

Convert the given temperature in Celsius to Kelvin: T=22.0+273.15=295.15 K. This allows the use of consistent units throughout the calculations.
04

Finding the Pressure-Volume Ratio

Calculate the volume ratio from work done: W=nRTln(V2V1)392=0.305×8.314×295.15×ln(V2V1). Solve for ln(V2V1), which provides ln(V2V1)0.058.
05

Solving for the Volume Ratio

From the previous step, ln(V2V1)=0.058 implies V2V1=e0.0581.06. Since P1V1=P2V2, then P1=P2V2V1.
06

Calculating Initial Pressure

Using the ratio V2V1=1.06 and P2=1.76 atm, find P1=1.76×1.061.87 atm.
07

Drawing the pV-Diagram

In a pV-diagram, plot the pressure on the y-axis and volume on the x-axis. The curve will show a hyperbolic shape (since the product PV is constant), starting from a higher volume and lower pressure (P1) to a lower volume and higher pressure (P2).

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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 equation in thermodynamics that describes the behavior of ideal gases. It is represented by the formula: PV=nRT, where:
  • P is the pressure of the gas
  • V is the volume of the gas
  • n is the number of moles of the gas
  • R is the ideal gas constant, approximately 8.314 J/mol·K
  • T is the temperature in Kelvin
This law assumes that the gas particles do not interact and that the volume of the gas particles is negligible compared to the volume of the container. It is especially useful for predicting the behavior of gases under various conditions of pressure, volume, and temperature. When applying the Ideal Gas Law, it is crucial to ensure that all units are consistent, particularly the temperature which must be in Kelvin. This law applies very well to many common gases at low pressure and high temperature.
Isothermal Process
An isothermal process is a thermodynamic process that occurs at a constant temperature. This means that the heat transferred into or out of the system adjusts to keep the temperature constant. For ideal gases, there are some important implications:
  • The internal energy of the gas remains unchanged because internal energy depends on temperature.
  • The work done by or on the gas equals the heat exchanged during the process; in symbols, Q=W.
  • The equation P1V1=P2V2 holds true, demonstrating that the product of pressure and volume remains constant during the process.
During an isothermal compression, like the one presented in the exercise, the gas is compressed by the surroundings doing work on it, while any heat produced is simultaneously transferred out of the gas to maintain constant temperature. Such processes are graphically represented by a hyperbolic curve on a Pressure-Volume diagram, indicating that as the volume decreases, the pressure increases.
Work Done on Gas
Work done on a gas during an isothermal process is an important concept in thermodynamics. During such a process, the total work done can be calculated by:W=nRTln(V2V1)where:
  • W is the work done
  • n is the number of moles of gas
  • R is the gas constant
  • T is the temperature held constant in Kelvin
  • V1 and V2 are the initial and final volumes respectively
It shows that work done on the gas depends on the ratio of final volume to initial volume. When calculating the work done, the sign convention can be important: work done on the gas is considered positive, while work done by the gas is considered negative. Understanding these principles helps in analyzing various thermodynamic processes and predicting how a system will respond under isothermal conditions.
Pressure-Volume Diagram
A Pressure-Volume (PV) diagram is a graphical representation of the relationship between the pressure and volume of a gas. In these diagrams:
  • The y-axis typically represents pressure.
  • The x-axis represents volume.
  • Each point on the curve represents a different state of the gas during the process.
For an isothermal process involving an ideal gas, the PV diagram displays a hyperbolic curve. This curve reflects the fact that during isothermal compression (or expansion), the product PV stays constant: as the volume decreases, pressure increases proportionately.To sketch a PV diagram for an isothermal process like the one in the exercise:
  • Start at the initial pressure P1 and volume V1.
  • The curve will slope downwards towards the final pressure P2 and volume V2.
  • The curve follows a hyperbolic shape due to the inverse relationship between pressure and volume when temperature remains constant.
By analyzing these diagrams, we gain insight into how gases behave under different thermodynamic conditions, which is crucial for designing and evaluating engineering systems.

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Most popular questions from this chapter

A certain ideal gas has molar heat capacity at constant volume CV . A sample of this gas initially occupies a volume V0 at pressure p0 and absolute temperature T0 . The gas expands isobarically to a volume 2V0 and then expands further adiabatically to a final volume 4V0 . (a) Draw a pV-diagram for this sequence of processes. (b) Compute the total work done by the gas for this sequence of processes. (c) Find the final temperature of the gas. (d) Find the absolute value of the total heat flow Q into or out of the gas for this sequence of processes, and state the direction of heat flow.

During an isothermal compression of an ideal gas, 410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at constant volume, adding 1.36 × 104 J of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. Calculate (a) the final temperature of the gas; (b) the amount of work done by the gas; (c) the amount of heat added to the gas while it was expanding; (d) the change in internal energy of the gas for the whole process.

A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at 1.00 × 105 Pa and 300 K and occupies a volume of 1.50 L. The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature at 300 K. This continues until the pressure reaches 2.50 × 104 Pa. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal. (a) In a pV-diagram, show both processes. (b) Find the volume of the gas at the end of the first process, and the pressure and temperature at the end of the second process. (c) Find the total work done by the gas during both processes. (d) What would you have to do to the gas to return it to its original pressure and temperature?

Three moles of an ideal monatomic gas expands at a constant pressure of 2.50 atm; the volume of the gas changes from 3.20 × 102 m3 to 4.50 × 102 m3. Calculate (a) the initial and final temperatures of the gas; (b) the amount of work the gas does in expanding; (c) the amount of heat added to the gas; (d) the change in internal energy of the gas.

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