/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A tank contains 111.0 g chlorine... [FREE SOLUTION] | 91Ó°ÊÓ

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A tank contains 111.0 g chlorine gas \(\left(\mathrm{Cl}_{2}\right)\), which is at temperature \(82.0^{\circ} \mathrm{C}\) and absolute pressure \(5.70 \times 10^{5} \mathrm{Pa} .\) The temperature of the air outside the tank is \(20.0^{\circ} \mathrm{C}\). The molar mass of \(\mathrm{Cl}_{2}\) is \(70.9 \mathrm{g} / \mathrm{mol}\). (a) What is the volume of the tank? (b) What is the internal energy of the gas? (c) What is the work done by the gas if the temperature and pressure inside the tank drop to \(31.0^{\circ} \mathrm{C}\) and \(3.80 \times 10^{5} \mathrm{Pa},\) respectively, due to a leak?

Short Answer

Expert verified
The initial volume of the tank is \(V = 0.0306 \ m^3\), the internal energy of the gas is \(U = 7755.5 \ J\), and the work done by the gas due to the leak is \(W = -5700 \ J\).

Step by step solution

01

Convert mass of chlorine gas to moles

First, we need to convert the mass of chlorine gas to moles using the molar mass formula: n = mass / molar mass. Given, the mass of Cl2 is 111.0 g, and the molar mass of Cl2 is 70.9 g/mol. So, we have: n = 111.0 g / 70.9 g/mol = 1.565 moles.
02

Find the initial volume of the tank

To find the initial volume of the tank, we can use the ideal gas law formula: PV = nRT. Given P = 5.70 x 10^5 Pa, n = 1.565 moles, R (ideal gas constant) = 8.314 J/mol.K, and T = 82°C + 273.15 = 355.15 K, we have: V = nRT / P = (1.565 mol)(8.314 J/mol.K)(355.15 K) / (5.70 x 10^5 Pa) = 0.0306 m^3.
03

Calculate the internal energy of the gas

To find the internal energy of the gas, we use the relation between internal energy and temperature for ideal gases: U = (3/2)nRT. Using the values of n, R, and T from Step 1 and Step 2, we get: U = (3/2)(1.565 mol)(8.314 J/mol.K)(355.15 K) = 7755.5 J.
04

Find the final volume of the tank after the leak

After the leak, the temperature and pressure inside the tank drop to 31°C and 3.80 x 10^5 Pa, respectively. We need to find the new volume of the tank using the ideal gas law formula: PV = nRT. T = 31°C + 273.15 = 304.15 K, P = 3.80 x 10^5 Pa, and n = 1.565 moles. So, we have: V = nRT / P = (1.565 mol)(8.314 J/mol.K)(304.15 K) / (3.80 x 10^5 Pa) = 0.0426 m^3.
05

Calculate the work done by the gas

Now, we will calculate the work done by the gas using the pressure-volume work formula: W = -P(ΔV). ΔV = Vfinal - Vinitial = 0.0426 m^3 - 0.0306 m^3 = 0.0120 m^3. To find the average pressure (P_avg), we can use the following formula: P_avg = (P_initial + P_final) / 2 = (5.70 x 10^5 Pa + 3.80 x 10^5 Pa) / 2 = 4.75 x 10^5 Pa. Now we can calculate the work done by the gas: W = -P_avg(ΔV) = -(4.75 x 10^5 Pa)(0.0120 m^3) = -5700 J. From the given data and steps, the results are: (a) The initial volume of the tank: V = 0.0306 m^3 (b) The internal energy of the gas: U = 7755.5 J (c) The work done by the gas: W = -5700 J.

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Key Concepts

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

Chlorine Gas Properties
Chlorine gas, denoted as \( \text{Cl}_2 \), is a diatomic molecule, meaning it consists of two atoms of chlorine bonded together. Unlike some other gases, it is yellowish-green and has a distinctive, pungent odor. Chlorine is heavier than air, and it is known for its reactivity, which makes it a useful molecule in chemical reactions and industrial applications. Understanding the properties of chlorine gas is crucial when working with it, as it can be hazardous to health. When studying chlorine gas, it is important to keep in mind:
  • The molecular weight is 70.9 g/mol, which impacts calculations involving moles.
  • Temperature and pressure conditions greatly influence its behavior according to the ideal gas law.
  • It is stored under pressure in tanks, requiring careful handling to ensure safety.
Internal Energy Calculation
In thermodynamics, the internal energy of a gas refers to the total energy contained within the system due to both the random motion of atoms and molecules and the potential energy of particles when they interact. For ideal gases, the internal energy is directly related to temperature. This relationship can be quantified using the equation:\[ U = \left( \frac{3}{2} \right)nRT \]where \( U \) is the internal energy, \( n \) represents the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.While dealing with gases like chlorine:
  • The internal energy is dependent on the kinetic activity of the molecules, which is affected by temperature changes.
  • For monoatomic ideal gases, internal energy focuses on translational motion, but for diatomic molecules like chlorine, rotational energy also contributes.
  • This formula's simplicity makes it an ideal starting point for calculations, provided the gas behaves close to ideal conditions.
Pressure-Volume Work
Pressure-volume work is a type of work done when there is an expansion or compression of gas within a system. This term is crucial when determining how work interacts with gases in enclosed spaces, such as tanks. Whenever a gas changes its volume, due to temperature or pressure fluctuations, work is either done on or by the gas.The work done by gas in such systems can be calculated using the formula:\[ W = -P_{\text{avg}}(\Delta V) \]where \( W \) is the work done, \( P_{\text{avg}} \) is the average pressure, and \( \Delta V \) signifies the change in volume.Key points to consider:
  • Ensure conditions are known for initial and final states to calculate changes accurately.
  • Always take the average pressure, especially if pressure changes during the process.
  • The negative sign indicates that work is done by the system when the volume increases.
Moles Conversion
Conversion of mass to moles is a foundational concept in chemistry and is essential for working with the ideal gas law. To calculate moles from mass, the molecular weight or molar mass of the substance is needed. Chlorine, whose molar mass is 70.9 g/mol, allows us to convert easily between units since:\[ n = \frac{\text{mass}}{\text{molar mass}} \]Here, \( n \) represents the number of moles. This calculation is critical when using formulas that require input in moles, ensuring correct stoichiometry and energy computations.When practicing this concept:
  • Double-check molecular weights from reliable sources for accuracy.
  • Pay attention to unit conversions to maintain consistency in calculations.
  • This conversion step is crucial before plugging values into any gas law equations like PV = nRT.

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Most popular questions from this chapter

The gasoline internal combustion engine operates in a cycle consisting of six parts. Four of these parts involve, among other things, friction, heat exchange through finite temperature differences, and accelerations of the piston; it is irreversible. Nevertheless, it is represented by the ideal reversible Otto cycle, which is illustrated below. The working substance of the cycle is assumed to be air. The six steps of the Otto cycle are as follows: i. Isobaric intake stroke (OA). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure \(p_{0}\) as the piston expands, increasing the volume of the cylinder from zero to \(V_{A}\) ii. Adiabatic compression stroke \((A B) .\) The temperature of the mixture rises as the piston compresses it adiabatically from a volume \(V_{\mathrm{A}}\) to \(V_{\mathrm{B}}\) iii. Ignition at constant volume (BC). The mixture is ignited by a spark. The combustion happens so fast that there is essentially no motion of the piston. During this process, the added heat \(Q_{1}\) causes the pressure to increase from \(p_{B}\) to \(p_{C}\) at the constant volume \(V_{\mathrm{B}}\left(=V_{\mathrm{C}}\right)\) iv. Adiabatic expansion (CD). The heated mixture of gasoline and air expands against the piston, increasing the volume from \(V_{C}\) to \(V_{D}\) This is called the power stroke, as it is the part of the cycle that delivers most of the power to the crankshaft. v. Constant-volume exhaust \((D A)\). When the exhaust valve opens, some of the combustion products escape. There is almost no movement of the piston during this part of the cycle, so the volume remains constant at \(V_{A}\left(=V_{D}\right)\) Most of the available energy is lost here, as represented by the heat exhaust \(Q_{2}\) vi. Isobaric compression (AO). The exhaust valve remains open, and the compression from \(V_{A}\) to zero drives out the remaining combustion products. (a) Using (i) \(e=W / Q_{1} ;\) (ii) \(W=Q_{1}-Q_{2} ;\) and (iii) \(Q_{1}=n C_{\nu}\left(T_{C}-T_{B}\right), Q_{2}=n C_{\nu}\left(T_{D}-T_{A}\right),\) show that \(e=1-\frac{T_{D}-T_{A}}{T_{C}-T_{B}}\) (b) Use the fact that steps (ii) and (iv) are adiabatic to show that \(e=1-\frac{1}{r^{\gamma-1}}\) where \(r=V_{A} / V_{B}\) The quantity \(r\) is called the compression ratio of the engine. (c) In practice, \(r\) is kept less than around 7 . For larger values, the gasoline-air mixture is compressed to temperatures so high that it explodes before the finely timed spark is delivered. This preignition causes engine knock and loss of power. Show that for \(r=6\) and \(\gamma=1.4\) (the value for air), \(e=0.51,\) or an efficiency of \(51 \%\) Because of the many irreversible processes, an actual internal combustion engine has an efficiency much less than this ideal value. A typical efficiency for a tuned engine is about \(25 \%\) to \(30 \%\)

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