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As \(3.0\) liters of ideal gas at \(27^{\circ} \mathrm{C}\) is heated, it expands at a constant pressure of \(2.0\) atm. How much work is done by the gas as its temperature is changed from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The work done by the gas is approximately 405.3 J.

Step by step solution

01

Convert Temperature to Kelvin

First, convert the temperatures given in Celsius to Kelvin. We have the initial temperature \(T_1 = 27^{\circ} \text{C} = 27 + 273.15 = 300.15 \text{ K}\) and the final temperature \(T_2 = 227^{\circ} \text{C} = 227 + 273.15 = 500.15 \text{ K}\).
02

Apply the Ideal Gas Law

Using the ideal gas equation, which is \( PV = nRT \), remember that since the pressure \( P \) and the gas constant \( R \) are constant, the volume is directly proportional to temperature. Thus, \( V_2 / V_1 = T_2 / T_1 \).
03

Determine Final Volume

Given the initial volume \( V_1 = 3.0 \text{ L} \), calculate the final volume \( V_2 \): \[ V_2 = V_1 \times \frac{T_2}{T_1} = 3.0 \times \frac{500.15}{300.15} \approx 5.0 \text{ L}. \]
04

Calculate Work Done

The work done by the gas during expansion at constant pressure is given by \( W = P \Delta V \), where \( \Delta V = V_2 - V_1 \). Convert the pressure from atm to \( \text{Pa} \) (1 atm = 101325 Pa): \( P = 2.0 \times 101325 \). Calculate the work: \[ \Delta V = 5.0 - 3.0 = 2.0 \text{ L} = 0.002 \text{ m}^3 \] \[ W = 2.0 \times 101325 \times 0.002 = 405.3 \text{ J}. \]
05

Conclusion

Thus, the work done by the gas as it expands is approximately 405.3 joules.

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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 concept in thermodynamics. It's expressed as \( PV = nRT \), where:
  • \( P \) represents pressure
  • \( V \) is volume
  • \( n \) is the number of moles
  • \( R \) is the universal gas constant
  • \( T \) is the temperature in Kelvin
In this exercise, we are dealing with a situation where the pressure remains constant, allowing us to use the concept that volume (V) is directly proportional to temperature (T) when rearranging the equation. This relationship helps to determine how much the volume changes as the temperature of an ideal gas is increased or decreased at a constant pressure. By understanding this relationship, we can predict the behavior of the gas under different conditions. This law is crucial for solving problems involving gases because it connects macroscopic properties like pressure, temperature, and volume, making it possible to solve for one when the others are known.
Gas Expansion
Gas expansion is an important process in thermodynamics, particularly when dealing with ideal gases. It involves the increase in volume of a gas as its temperature rises or pressure decreases, or both. In our particular case, the gas expands because the temperature is increased while keeping pressure constant. For a constant pressure process, the gas expands more as the temperature increases. During this expansion, we are interested in how the final volume is affected by the change in temperature. With the assumption of ideal gas behavior, we can use the equation \( V_2 / V_1 = T_2 / T_1 \) to find the final volume \( V_2 \). This expression shows us that the final and initial volumes of the gas are directly proportional to their respective temperatures when the pressure is constant. Understanding how gases expand and contract under varying conditions is foundational in predicting and manipulating gas behavior in both scientific and industrial applications.
Temperature Conversion
Temperature conversion is a straightforward yet crucial step when working with thermodynamics and the ideal gas law. The rule of thumb is to always convert temperatures from Celsius to Kelvin before doing any calculations that involve the physical properties of gases. The Kelvin scale is preferred because it is an absolute temperature scale, which means it starts at absolute zero, where all molecular motion theoretically stops. This makes it ideal for calculations involving temperature relations.The conversion is simple:
  • To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.
In our exercise, we start with initial and final temperatures in Celsius: \(27^{\circ} \text{C}\) and \(227^{\circ} \text{C}\). Converting them gives us 300.15 K and 500.15 K respectively. Always remember to check units—using the wrong scale can lead to incorrect conclusions in your calculations.
Work Calculation
Work calculation in thermodynamics is an essential aspect, often related to energy transformations. In our exercise, we calculate the work done by an expanding gas at constant pressure using the formula \( W = P \Delta V \), where:
  • \( W \) is the work done by the gas.
  • \( P \) is the pressure, which must be converted to appropriate units (Pascals in our case).
  • \( \Delta V \) is the change in volume, found by subtracting the initial volume from the final volume.
Here the gas expands from 3.0 L to 5.0 L, so \( \Delta V \) is 2.0 L which we convert to cubic meters for calculation (0.002 m³). We then multiply by the pressure expressed in Pascals—2 atm equals 202650 Pa—for accurate work computation. Finally, the calculation yields a work done of 405.3 J. Understanding these calculations helps us grasp how mechanical work and thermal energy exchange are related in physical systems.

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Most popular questions from this chapter

How much work is done by an ideal gas in expanding isothermally from an initial volume of \(3.00\) liters at \(20.0\) atm to a final volume of \(24.0\) liters? For an isothermal expansion by an ideal gas, $$ \begin{aligned} \Delta W &=P_{1} V_{1} \ln \left(\frac{V_{2}}{V_{1}}\right) \\ &=\left(20.0 \times 1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(3.00 \times 10^{-3} \mathrm{~m}^{3}\right) \ln \left(\frac{24.0}{3.00}\right)=12.6 \mathrm{~kJ} \end{aligned} $$

Water is boiled at \(100^{\circ} \mathrm{C}\) and \(1.0\) atm. Under these conditions, 1.0 g of water occupies \(1.0 \mathrm{~cm}^{3}, 1.0\) g of steam occupies 1670 \(\mathrm{cm}^{3}\), and \(L_{v}=540 \mathrm{cal} / \mathrm{g}\). Find \((a)\) the external work done when \(1.0\) g of steam is formed at \(100^{\circ} \mathrm{C}\) and \((b)\) the increase in internal energy.

For each of the following adiabatic processes, find the change in internal energy. ( \(a\) ) A gas does \(5 \mathrm{~J}\) of work while expanding adiabatically. (b) During an adiabatic compression, \(80 \mathrm{~J}\) of work is done on a gas. During an adiabatic process, no heat is transferred to or from the system. (a) \(\Delta U=\Delta Q-\Delta W=0-5 \mathrm{~J}=-5 \mathrm{~J}\) (b) \(\Delta U=\Delta Q-\Delta W=0-(-80 \mathrm{~J})=+80 \mathrm{~J}\)

The temperature of \(5.00 \mathrm{~kg}\) of \(\mathrm{N}_{2}\) gas is raised from \(10.0{ }^{\circ} \mathrm{C}\) to \(130.0{ }^{\circ} \mathrm{C}\). If this is done at constant volume, find the increase in internal energy \(\Delta U\). Alternatively, if the same temperature change now occurs at constant pressure determine both \(\Delta V\) and the external work \(\Delta W\) done by the gas. For \(\mathrm{N}_{2}\) gas, \(c_{v}=0.177\) cal \(/ \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(c_{p}=0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). If the gas is heated at constant volume, then no work is done during the process. In that case \(\Delta W=0\), and the First Law tells us that \((\Delta Q)_{v}=\Delta U\). Because \((\Delta Q)_{v}=c_{v} m \Delta T\), \(\Delta U=(\Delta Q)_{v}=\left(0.177 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120^{\circ} \mathrm{C}\right)=106 \mathrm{kcal}=443 \mathrm{~kJ}\) The temperature change is a manifestation of the internal energy change. When the gas is heated \(120^{\circ} \mathrm{C}\) at constant pressure, the same change in internal energy occurs. In addition, however, work is done. The First Law then becomes $$(\Delta Q)_{\mathrm{p}}=\Delta U+\Delta W=443 \mathrm{~kJ}+\Delta W$$ But \((\Delta Q)_{\mathrm{p}}=c_{p} m \Delta T=\left(0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120^{\circ} \mathrm{C}\right)\) $$=149 \mathrm{kcal}=623 \mathrm{~kJ}$$ Hence \(\Delta W=(\Delta Q)_{\mathrm{p}}-\Delta U=623 \mathrm{~kJ}-443 \mathrm{~kJ}=180 \mathrm{~kJ}\)

Molecular oxygen having a mass of \(10.0\) g is in a cylinder sealed with a movable piston. The gas is heated from \(0.00^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\) at a constant pressure and expands. Given that \(c_{p}\) for \(\mathrm{O}_{2}\) is \(0.919\) \(\mathrm{kJ} / \mathrm{kg}\), how much heat was received by the gas?

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