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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
Draw a horizontal line on a p-V diagram; work done is 1321.24 J.

Step by step solution

01

Convert Temperatures to Kelvin

First, convert the given temperatures in Celsius to Kelvin, since calculations involving gases typically use Kelvin. The formula for conversion is: \[ T(K) = T(^{\circ}C) + 273.15 \] Thus, the initial temperature is \( T_1 = 27^{\circ}C + 273.15 = 300.15 \, K \) and the final temperature is \( T_2 = 107^{\circ}C + 273.15 = 380.15 \, K \).
02

Understand the Given Information

Recall that when a gas is heated at constant pressure, the volume changes. The process is an isobaric expansion. We know that the initial conditions involve 2 moles of the gas and that the pressure remains constant throughout.
03

Draw the p-V Diagram

In an isobaric process, the pressure \( p \) remains constant as the volume \( V \) changes. The \( p-V \) diagram will be a horizontal line, starting from an initial volume \( V_1 \) (at 300.15 K) to a larger volume \( V_2 \) (at 380.15 K). Label the vertical axis as pressure \( p \) and the horizontal axis as volume \( V \), with the line showing the heating process from the initial to the final volume.
04

Use Ideal Gas Law to Relate Volumes

Given that the pressure is constant, use the Ideal Gas Law \( pV = nRT \) to relate the changes in volume and temperature. Since \( p \) and \( n \) are constant, the relationship is \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). Let \( V_1 \) and \( V_2 \) be the initial and final volumes.
05

Calculate Work Done

The work done by the gas in an isobaric process is given by \( W = p(V_2 - V_1) \). Rearrange the Ideal Gas Law to express work in terms of temperature: \( W = nR(T_2 - T_1) \). Substitute \( n = 2 \) moles, \( R = 8.314 \, J/(mol \cdot K) \), \( T_1 = 300.15 \, K \), and \( T_2 = 380.15 \, K \). Hence, \[ W = 2 \times 8.314 \times (380.15 - 300.15) = 2 \times 8.314 \times 80 = 1321.24 \, J \].

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Key Concepts

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

Isobaric Process
An isobaric process is one where the pressure remains constant throughout the entire reaction or change. This is a common occurrence in thermodynamics when a gas changes its volume, but the external pressure doesn't vary. In our problem, as the gas is heated, its volume increases while maintaining a steady pressure. This means no spikes or drops in pressure occur as the temperature and volume change.
In an isobaric process, the relationship between volume and temperature is directly proportional. This means that if you increase the temperature, the volume increases too, assuming pressure and the amount of gas stay constant. The mathematical representation of an isobaric process is derived from the Ideal Gas Law, expressed as \(PV = nRT\), where \(P\), \(V\), and \(T\) are pressure, volume, and temperature respectively, \(n\) is the amount of substance (in moles), and \(R\) is the gas constant.
During this type of process, you will usually find a horizontal line on a PV diagram, which indicates a constant pressure with changes in volume.
Work Done by Gas
The concept of work done by a gas in an isobaric process is a crucial aspect of understanding thermodynamics. When gas does work, it expands against an external pressure. In our specific problem, this work can be calculated using the formula: \(W = p(V_2 - V_1)\).
The work done is positive since the gas is expanding, pushing against the pressure while it increases in volume. Alternatively, using the Ideal Gas Law and knowing that pressure is constant, we can express work in terms of temperature like this: \(W = nR(T_2 - T_1)\). Here, \(T_2\) and \(T_1\) are the final and initial temperatures in Kelvin, \(n\) is the number of moles, and \(R\) is the gas constant.
For our case, substituting in the values: the work done by the gas was \(1321.24\, J\). This represents the energy transferred as the gas expanded.
PV Diagram
A PV diagram is a graphical representation of the relationship between pressure (P) and volume (V) of a gas. This tool is incredibly handy for visualizing processes like the one described in our problem. Since we are dealing with an isobaric process, the PV diagram in this case is simple: it's a horizontal line.
The horizontal line indicates constant pressure while the volume changes. This line starts at the initial volume when the gas is at the initial given temperature of 300.15 K and ends at the final volume when the gas has reached 380.15 K. The pressure axis is marked vertically, and the volume is horizontal. It's very important to clearly label each axis to better comprehend the changes taking place.
PV diagrams can also be useful in other thermodynamic processes like isochoric, isothermal, and adiabatic processes, each having its unique plot.
Temperature Conversion
Understanding temperature conversion is fundamental in thermodynamics as it allows us to move between Celsius and Kelvin scales with ease. Kelvin is the SI unit for temperature used in most thermodynamic equations. The conversion formula is quite simple: add 273.15 to the Celsius temperature to get Kelvin.
In our exercise, the problem specified temperatures in Celsius. By converting these temperatures to Kelvin using the formula \( T(K) = T(^{\circ}C) + 273.15 \), we found that the initial temperature was 300.15 K and the final temperature was 380.15 K.
This conversion is crucial because the Ideal Gas Law, which governs the behavior of gases in processes like the one we are examining, requires temperature to be in Kelvin to ensure accurate calculations. This step provides a crucial bridge between everyday experiences of temperature and the precise calculations needed in scientific contexts.

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Most popular questions from this chapter

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