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Chapter 20: Problem 19

Three kilomoles (6.00 kg) of hydrogen gas at S.T.P. expands isobarically to precisely twice its volume. (a) What is the final temperature of the gas? (b) What is the expansion work done by the gas? ( \(c\) ) By how much does the internal energy of the gas change? ( \(d\) ) How much heat enters the gas during the expansion? For \(\mathrm{H}_{2}, c_{v}=\) \(10.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Assume the hydrogen will behave as an ideal gas. (a) From \(P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}\) with \(P_{1}=P_{2}\), $$T_{2}=T_{1}\left(\frac{V_{2}}{V_{1}}\right)=(273 \mathrm{~K})(2.00)=546 \mathrm{~K}$$ (b) Because 1 kmol at S.T.P. occupies \(22.4 \mathrm{~m}^{3}\), we have \(V_{1}=\) \(67.2 \mathrm{~m}^{3}\). Then $$\Delta W=P \Delta V=P\left(V_{2}-V_{1}\right)=\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(67.2 \mathrm{~m}^{3}\right)=6.8 \mathrm{MJ}$$ (c) To raise the temperature of this ideal gas by \(273 \mathrm{~K}\) at constant volume requires $$\Delta Q=c_{v} m \Delta T=(10.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(6.00 \mathrm{~kg})(273 \mathrm{~K})=16.4 \mathrm{MJ}$$ Because the volume is constant here, no work is done and \(\Delta Q\) equals the internal energy that must be added to the \(6.00 \mathrm{~kg}\) of \(\mathrm{H}_{2}\) to change its temperature from \(273 \mathrm{~K}\) to \(546 \mathrm{~K}\). Therefore, \(\Delta U=16.4 \mathrm{MJ} .\) ( \(d\) ) The system obeys the First Law during the process and so $$ \Delta Q=\Delta U+\Delta W=16.4 \mathrm{MJ}+6.8 \mathrm{MJ}=23.2 \mathrm{MJ} $$

Short Answer

Expert verified
(a) 546 K, (b) 6.8 MJ, (c) 16.4 MJ, (d) 23.2 MJ.

Step by step solution

01

Initial Assumptions and Formulas

We are given 3 kilomoles (or 6 kg) of hydrogen gas at Standard Temperature and Pressure (S.T.P.). The gas expands isobarically to twice its original volume, and we need to find the final temperature, the work done, the change in internal energy, and the amount of heat entering the gas. We'll assume hydrogen behaves as an ideal gas.
02

Calculate Final Temperature

Using the formula for the isobaric process, \( P_1 V_1 / T_1 = P_2 V_2 / T_2 \) and knowing that \( P_1 = P_2 \) and \( V_2 = 2V_1 \), we can solve for the final temperature \( T_2 \). Since \( T_1 = 273 \text{ K} \) at S.T.P., the final temperature \( T_2 \) is given by: \[ T_2 = T_1 \left( \frac{V_2}{V_1} \right) = 273 \text{ K} \times 2 = 546 \text{ K}. \]
03

Calculate Work Done by the Gas

In an isobaric expansion, the work done \( \Delta W \) is calculated by \( P \Delta V = P (V_2 - V_1) \). Since 1 kilomole of hydrogen occupies 22.4 m鲁 at S.T.P., 3 kilomoles occupy 67.2 m鲁 (\( V_1 \)). So \( V_2 = 2 \times 67.2 = 134.4 \text{ m}^3 \). Thus, \( \Delta W = 1.01 \times 10^5 \text{ N/m}^2 \times (134.4 - 67.2) \text{ m}^3 = 6.8 \text{ MJ}. \)
04

Calculate Change in Internal Energy

The change in internal energy \( \Delta U \) is calculated as \( \Delta U = c_v m \Delta T \). Here, \( \Delta T = T_2 - T_1 = 546 \text{ K} - 273 \text{ K} = 273 \text{ K} \). Substituting the given values, \( c_v = 10.0 \text{ kJ/kg} \cdot \text{K} \) and \( m = 6 \text{ kg} \), gives: \[ \Delta U = 10.0 \text{ kJ/kg} \cdot \text{K} \times 6 \text{ kg} \times 273 \text{ K} = 16.4 \text{ MJ}. \]
05

Calculate Heat Entering the Gas

Using the First Law of Thermodynamics, \( \Delta Q = \Delta U + \Delta W \), we find the heat entering the gas. With \( \Delta U = 16.4 \text{ MJ} \) and \( \Delta W = 6.8 \text{ MJ} \), the total heat is: \[ \Delta Q = 16.4 \text{ MJ} + 6.8 \text{ MJ} = 23.2 \text{ MJ}. \]

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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 a transformation in which the pressure of the gas remains constant. This means that throughout the process, the initial and final pressures are equal. Isobaric processes are common in thermodynamics, especially involving ideal gases in a piston system, where the pressure against the piston remains constant while volume changes.
In the exercise you encountered, the hydrogen gas expands isobarically, doubling its volume from its original state. To examine such a process, we use the isobaric condition, represented by the ideal gas law in the form \[ P_1 V_1 / T_1 = P_2 V_2 / T_2 \] where \( P_1 = P_2 \) because the process is isobaric. This equation helps determine the relationship between the initial and final states of the gas, allowing us to calculate the final temperature \( T_2 \).
This calculation is straightforward because an isobaric process simplifies many relationships, focusing mainly on how volume and temperature relate when pressure does not change.
Internal Energy Change
Internal energy is the total energy contained within a system, which includes kinetic and potential energies at a microscopic level. For an ideal gas, changes in internal energy are directly related to changes in temperature because potential energies like interatomic forces are considered negligible.
The change in internal energy \( \Delta U \) can be calculated using the formula \[ \Delta U = c_v m \Delta T \]where \( c_v \) is the specific heat at constant volume, \( m \) is the mass of the gas, and \( \Delta T \) is the change in temperature. In this exercise, since the gas temperature increases from \( 273 \text{ K} \) to \( 546 \text{ K} \), the internal energy increases accordingly.
Internal energy change is crucial to understand because it provides insight into how much energy the system needs to change state and how this energy interacts with other variables like heat and work.
Expansion Work
During an isobaric expansion of a gas, the system performs expansion work. Work in thermodynamics is the energy transferred when an object is moved over a distance by an external force. In the context of a gas, expansion work \( \Delta W \) involves a gas pushing against a boundary such as a piston, causing the volume to change.
The formula to calculate the work done during an isobaric process is \[ \Delta W = P \Delta V = P (V_2 - V_1) \]where \( P \) is the constant pressure, and \( \Delta V \) is the change in volume of the gas. In our exercise, this calculation involves the initial and final volumes of the gas, illustrating the energy required to expand the gas under constant pressure.
Understanding expansion work helps explain how energy is transferred to or from a system as it changes volumes, such as performing mechanical work during expansion.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that the energy within a closed system must be conserved. This means energy cannot be created or destroyed, only transferred or transformed.
Mathematically, this principle is represented as \[ \Delta Q = \Delta U + \Delta W \]where \( \Delta Q \) is the heat added to the system, \( \Delta U \) is the change in internal energy, and \( \Delta W \) is the work done by the system. In our exercise, you calculated the amount of heat entering the hydrogen gas through this formula.
This concept is pivotal in thermodynamics because it links heat exchange, work done, and changes in internal energy, providing a framework to analyze processes and energy flows within any thermodynamic system.

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Most popular questions from this chapter

The specific heat of water is \(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). By how many joules does the internal energy of \(50 \mathrm{~g}\) of water change as it is heated from \(21^{\circ} \mathrm{C}\) to \(37^{\circ} \mathrm{C}\) ? Assume that the expansion of the water is negligible. The heat added to raise the temperature of the water is \(\Delta Q=c m \Delta T=(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.050 \mathrm{~kg})\left(16^{\circ} \mathrm{C}\right)=3.4 \times 10^{3} \mathrm{~J}\) Notice that \(\Delta T\) in Celsius is equal to \(\Delta T\) in kelvin. If we ignore the slight expansion of the water, no work was done on the surroundings and so \(\Delta W=0\). Then, the First Law, \(\Delta Q=\Delta U+\Delta W\), tells us that $$ \Delta U=\Delta Q=3.4 \mathrm{~kJ} $$

The specific heat of air at constant volume is \(0.175 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} .(a)\) By how much does the internal energy of \(5.0\) g of air change as it is heated from \(20^{\circ} \mathrm{C}\) to \(400{ }^{\circ} \mathrm{C} ?(b)\) Suppose that \(5.0\) g of air is adiabatically compressed so as to rise its temperature from \(20^{\circ} \mathrm{C}\) to \(400{ }^{\circ} \mathrm{C}\). How much work must be done on the air to compress it?

Twenty cubic centimeters of monatomic gas at \(12{ }^{\circ} \mathrm{C}\) and 100 \(\mathrm{kPa}\) is suddenly (and adiabatically) compressed to \(0.50 \mathrm{~cm}^{3}\). Assume that we are dealing with an ideal gas. What are its new pressure and temperature? For an adiabatic change involving an ideal gas, \(P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}\) where \(\mathrm{Y}=1.67\) for a monatomic gas. Hence, $$P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(1.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{20}{0.50}\right)^{1.67}=4.74 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=47 \mathrm{MPa}$$ To find the final temperature, we could use \(P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}\). Instead, let us use $$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma}$$ or $$T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=(285 \mathrm{~K})\left(\frac{20}{0.50}\right)^{0.67}=(285 \mathrm{~K})(11.8)=3.4 \times 10^{3} \mathrm{~K}$$ As a check, $$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$$ $$ \begin{aligned} \frac{\left(1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(20 \mathrm{~cm}^{3}\right)}{285 \mathrm{~K}} &=\frac{\left(4.74 \times 107 \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.50 \mathrm{~cm}^{3}\right)}{3370 \mathrm{~K}} \\ 7000 &=7000 \end{aligned} $$

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

Find \(\Delta W\) and \(\Delta U\) for a \(6.0\) -cm cube of iron as it is heated from 20 \({ }^{\circ} \mathrm{C}\) to \(300{ }^{\circ} \mathrm{C}\) at atmospheric pressure. For iron, \(c=0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and the volume coefficient of thermal expansion is \(3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}\) \- 1 . The mass of the cube is 1700 g. Given that \(\Delta T=300^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=280^{\circ} \mathrm{C}\), \(\Delta Q=c m \Delta T=\left(0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(1700 \mathrm{~g})\left(280^{\circ} \mathrm{C}\right)=52 \mathrm{kcal}\) To find that the work done by the expansion of the cube, we need to determine \(\Delta V\). The volume of the cube is \(V=(6.0 \mathrm{~cm})^{3}=216 \mathrm{~cm}^{3} .\) Using \((\Delta V) / V\) \(=\beta \Delta T\), $$\begin{array}{l} \Delta V=V \beta \Delta T=\left(216 \times 10^{-6} \mathrm{~m}^{3}\right)\left(3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(280^{\circ} \mathrm{C}\right)=2.18 \times 10^{-6} \mathrm{~m}^{3} \end{array}$$ But the First Law tells us that $$\begin{aligned} \Delta U &=\Delta Q-\Delta W=(52000 \mathrm{cal})(4.184 \mathrm{~J} / \mathrm{cal})-0.22 \mathrm{~J} \\ &=218000 \mathrm{~J}-0.22 \mathrm{~J} \approx 2.2 \times 10^{5} \mathrm{~J} \end{aligned}$$ Notice how very small the work of expansion against the atmosphere is in comparison to \(\Delta U\) and \(\Delta Q .\) Often \(\Delta W\) can be neglected when dealing with liquids and solids.
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