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Chapter 20: Problem 12

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, $$ \begin{aligned} \Delta W &=P_{1} V_{1} \ln \left(\frac{V_{2}}{V_{1}}\right) \\ &=\left(20.0 \times 1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(3.00 \times 10^{-3} \mathrm{~m}^{3}\right) \ln \left(\frac{24.0}{3.00}\right)=12.6 \mathrm{~kJ} \end{aligned} $$

Short Answer

Expert verified
The work done is 12.6 kJ.

Step by step solution

01

Identify the variables

First, identify the initial and final volumes and the initial pressure of the gas. These are given as: \(V_1 = 3.00\) liters, \(V_2 = 24.0\) liters, and \(P_1 = 20.0\) atm. Remember to convert volumes to cubic meters and pressure to pascals.
02

Convert units

Convert the initial volume from liters to cubic meters: \(V_1 = 3.00\) liters = \(3.00 \times 10^{-3}\) m鲁. Convert the final volume similarly: \(V_2 = 24.0\) liters = \(24.0 \times 10^{-3}\) m鲁. Convert the pressure from atmospheres to pascals: \(P_1 = 20.0\) atm = \(20.0 \times 1.01 \times 10^{5}\) N/m虏.
03

Use the work formula for isothermal expansion

Apply the formula for the work done during an isothermal expansion of an ideal gas: \( W = P_1 V_1 \ln \left(\frac{V_2}{V_1}\right)\). Substitute the converted values into this formula.
04

Calculate the natural logarithm

Calculate the natural logarithm part of the equation: \( \ln \left(\frac{V_2}{V_1}\right) = \ln \left(\frac{24.0}{3.00}\right) = \ln(8)\).
05

Perform the calculation

Compute the work done: \[ W = (20.0 \times 1.01 \times 10^{5} \text{ N/m}^2) \times (3.00 \times 10^{-3} \text{ m}^3) \times \ln(8) \]. After calculating, you find that \( W = 12.6 \text{ kJ} \).

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Key Concepts

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

Ideal Gas
An ideal gas is a theoretical concept in physics and chemistry. It represents a hypothetical gas made up of identical, non-interacting particles. The ideal gas follows a simple set of laws and makes problem-solving much easier.
Ideal gases are described by the ideal gas law, which states:
\[ PV = nRT \] where:- \(P\) is the pressure,- \(V\) is the volume,- \(n\) is the number of moles,- \(R\) is the ideal gas constant,- \(T\) is the temperature.
In our problem, the gas is expanding isothermally. This means it expands at a constant temperature. For ideal gases, this involves straightforward calculations once you know the initial and final conditions.
Work Calculation
Calculating work done during an isothermal expansion requires a specific formula used for ideal gases. The work done, \(W\), is given by:
\[ W = P_1 V_1 \, \ln \left(\frac{V_2}{V_1}\right) \] This formula highlights:- \(P_1\), the initial pressure.- \(V_1\), the initial volume.- \(V_2\), the final volume.
These elements come together in the equation to give us the work done, which measures energy transferred during the expansion. This process can be visualized as the area under a pressure-volume curve where the line is a hyperbola describing constant temperature.
Unit Conversion
Unit conversion is a crucial step in many physics problems to ensure all quantities are compatible.
For this problem:- Pressure needs to be converted from atmospheres to pascals: - 1 atm = \(1.01 \times 10^5\) N/m虏. - Therefore, \(20.0\) atm = \(20.0 \times 1.01 \times 10^5\) N/m虏.- Volumes need to be converted from liters to cubic meters: - 1 liter = \(10^{-3}\) m鲁. - Thus, \(3.00\) L = \(3.00 \times 10^{-3}\) m鲁 and \(24.0\) L = \(24.0 \times 10^{-3}\) m鲁.
Converting these units ensures calculations remain consistent and accurate.
Natural Logarithm
In this context, the natural logarithm function, denoted as \(\ln\), plays a significant role in determining the work done during isothermal expansion. Natural logarithms involve the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828.
The natural logarithm appears in the formula:
\[ \ln \left(\frac{V_2}{V_1}\right) \] To better understand, it calculates the ratio of final volume \(V_2\) to initial volume \(V_1\). Specifically, compute: - \(\ln(\frac{24.0}{3.00}) = \ln(8)\), a crucial step in finding the work output.
This function helps in accurately modeling the exponential nature of isothermal processes for an ideal gas.

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Most popular questions from this chapter

A \(2.0\) kg metal block \(\left(c=0.137 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) is heated from \(15^{\circ} \mathrm{C}\) to \(90{ }^{\circ} \mathrm{C}\). By how much does its internal energy change?

A spring \((k=500 \mathrm{~N} / \mathrm{m})\) supports a 400 -g mass, which is immersed in 900 g of water. The specific heat of the mass is \(450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The spring is now stretched \(15 \mathrm{~cm}\), and after thermal equilibrium is reached, the mass is released so it vibrates up and down. By how much has the temperature of the water changed when the vibration has stopped? The energy stored in the spring is dissipated by the effects of friction and goes to heat the water and mass. The energy stored in the stretched spring was $$\mathrm{PE}_{e}=\frac{1}{2} k x^{2}=\frac{1}{2}(500 \mathrm{~N} / \mathrm{m})(0.15 \mathrm{~m})^{2}=5.625 \mathrm{~J}$$ This energy appears as thermal energy that flows into the water and the mass. Using \(\Delta Q=\mathrm{cm} \Delta T\), $$5.625 \mathrm{~J}=(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.900 \mathrm{~kg}) \Delta T+(450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.40 \mathrm{~kg}) \Delta 7$$ which leads to $$ \Delta T=\frac{5.625 \mathrm{~J}}{3950 \mathrm{~J} / \mathrm{K}}=0.0014 \mathrm{~K} $$

What is the net work output per cycle for the thermodynamic cycle in Fig. \(20-4\) ? We know that the net work output per cycle is the area enclosed by the \(P-V\) diagram. We estimate that in area \(A B C A\) there are 22 squares, each of area $$\left(0.5 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.1 \mathrm{~m}^{3}\right)=5 \mathrm{~kJ}$$ Therefore, Area enclosed by cycle \(\approx(22)(5 \mathrm{~kJ})=1 \times 10^{2} \mathrm{~kJ}\) The net work output per cycle is \(1 \times 10^{2} \mathrm{~kJ}\).

How much does the internal energy of \(5.0\) g of ice at precisely \(0{ }^{\circ} \mathrm{C}\) increase as it is changed to water at \(0{ }^{\circ} \mathrm{C}\) ? Neglect the change in volume. The heat needed to melt the ice is $$\Delta Q=m L_{f}=(5.0 \mathrm{~g})(80 \mathrm{cal} / \mathrm{g})=400 \mathrm{cal}$$ No external work is done by the ice as it melts and so \(\Delta W=0\). Therefore, the First Law, \(\Delta Q=\Delta U+\Delta W\), tells us that $$ \Delta U=\Delta Q=(400 \mathrm{cal})(4.184 \mathrm{~J} / \mathrm{cal})=1.7 \mathrm{~kJ} $$

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 \(227^{\circ} \mathrm{C}\) ?
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