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A metal tank with volume 3.10 L will burst if the absolute pressure of the gas it contains exceeds 100 atm. (a) If 11.0 mol of an ideal gas is put into the tank at a temperature of \(23.0^{\circ} \mathrm{C},\) to what temperature can the gas be warmed before the tank ruptures? You can ignore the thermal expansion of the tank. (b) Based on your answer to part (a), is it reasonable to ignore the thermal expansion of the tank? Explain.

Short Answer

Expert verified
The gas can be warmed to approximately 72.77°C before the tank ruptures. Ignoring thermal expansion is reasonable due to the small temperature difference.

Step by step solution

01

Understand the Known Values

First, identify what's given in the problem:- Volume of the tank, \( V = 3.10 \) L (convert to m³: \( 3.10 \times 10^{-3} \) m³)- Number of moles, \( n = 11.0 \) mol- Initial temperature, \( T_1 = 23.0^{\circ} \text{C} = 296.15 \) K (Convert Celsius to Kelvin)- Maximum pressure before rupture, \( P_2 = 100 \) atm (convert to Pascals: \( 100 \times 1.01325 \times 10^5 \) Pa) The Universal gas constant, \( R = 8.314 \) J/(mol K).
02

Use the Ideal Gas Law Equation

Use the ideal gas law to state that at the initial condition, \( P_1V = nRT_1 \). We need to find \( P_1 \), the initial pressure:\[P_1V = nRT_1 \]Substitute the known values:\[P_1 = \frac{nRT_1}{V} = \frac{11.0 \times 8.314 \times 296.15}{3.10 \times 10^{-3}} \]
03

Calculate Initial Pressure

Calculate \( P_1 \) using the values from Step 2:\[P_1 = \frac{11.0 \times 8.314 \times 296.15}{3.10 \times 10^{-3}} \approx 8.75 \times 10^6 \, \text{Pa} \]Now we have the initial pressure.
04

Determine the Final Temperature Before Rupture

From the initial condition and ideal gas law, we use the final condition at rupture pressure \( P_2 \):\[P_2 = \frac{nRT_2}{V}\]Using \( T_2 \), the temperature at which the tank ruptures, solve for \( T_2 \):\[T_2 = \frac{P_2 \cdot V}{nR} = \frac{(100 \times 1.01325 \times 10^5) \cdot (3.10 \times 10^{-3})}{11.0 \times 8.314}\]
05

Calculate Final Temperature

Perform the calculation for \( T_2 \):\[T_2 = \frac{100 \times 1.01325 \times 10^5 \times 3.10 \times 10^{-3}}{11.0 \times 8.314} \approx 345.92 \, \text{K} \]Convert back to Celsius:\[T_2 \approx 345.92 - 273.15 \approx 72.77^{\circ} \text{C}\]
06

Evaluate Thermal Expansion Consideration

Based on \( T_2 \approx 72.77^{\circ}\text{C}\), evaluate the assumption:Since the difference between the initial and final temperature is relatively small (around 50°C), the volume change due to thermal expansion of the tank is minimal, thus reasonable to ignore in the calculation.

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Key Concepts

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

Gas Pressure Calculations
When dealing with gases, pressure is a key factor to consider, particularly within a confined space like a tank. Pressure, in simple terms, is the force exerted by the gas particles when they collide with the walls of their container. For our problem, it's crucial to understand that the tank will burst if the pressure exceeds a certain limit, in this case, 100 atm.

The Ideal Gas Law, represented by the equation \( PV = nRT \), helps us understand the relationship between pressure \( P \), volume \( V \), and temperature \( T \), with \( n \) being the number of moles of the gas and \( R \) the universal gas constant. It's used to determine the conditions under which our gas operates without leading to a tank rupture. We calculate initial pressure \( P_1 \) using volume, moles, and initial temperature to ensure it doesn't exceed the rupture pressure during thermal adjustments.

In the solution, the Ideal Gas Law allowed us to calculate the initial pressure and see how much it changes with temperature. Remember to always keep an eye on the units; changing atm to Pascal or Celsius to Kelvin is a crucial step that can change results dramatically.
Thermal Expansion
Thermal expansion is a phenomenon where materials expand upon heating. For gases, this means increased motion and collisions at higher temperatures that can raise pressure in a confined tank. However, in our exercise, we are instructed to ignore the tank's thermal expansion.

Why disregard it? If you calculate the final temperature of the gas and compare it to the initial temperature, the difference is small enough (only about 50°C) to have a negligible impact on the tank itself in reality. Materials like steel, often used for such tanks, have a relatively low coefficient of expansion, indicating very little dimensional change over such a temperature range.

Therefore, in this context, we prioritize the gas properties and calculations, as the tank's expansion would not significantly alter the results of pressure at the bursting point. Understanding when and why certain aspects can be ignored helps simplify complex problems without losing accuracy in approximations.
Temperature Conversions
Temperature plays a crucial role in gas calculations. When using the Ideal Gas Law, all temperature values must be in Kelvin (K). This is because Kelvin starts at absolute zero, the theoretically lowest possible temperature where particles have minimal thermal motion, making calculations mathematically and physically consistent.

To convert Celsius to Kelvin, you use the simple formula:
  • \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \)
This conversion was vital in our solution, as the gas initially was at 23.0°C, which is 296.15 K. Similarly, after increasing the temperature to when it bursts the tank at 345.92 K, converting back to Celsius helps understand the practical difference.

Keep in mind: always convert to Kelvin before inserting temperatures into any gas law equations, to ensure accuracy and avoid errors derived from incorrect scales.

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Most popular questions from this chapter

The speed of propagation of a sound wave in air at \(27^{\circ} \mathrm{C}\) is about 350 \(\mathrm{m} / \mathrm{s}\) . Calculate, for comparison, (a) \(v_{\mathrm{ms}}\) for nitrogen molecules and (b) the rms value of \(v_{x}\) at this temperature. The molar mass of nitrogen \(\left(\mathrm{N}_{2}\right)\) is \(28.0 \mathrm{g} / \mathrm{mol} .\)

(a) Calculate the total rotational kinetic energy of the molecules in 1.00 mol of a diatomic gas at 300 \(\mathrm{K}\) . (b) Calculate the moment of inertia of an oxygen molecule \(\left(\mathrm{O}_{2}\right)\) for rotation about either the \(y\) - or \(z\) -axis shown in Fig. 18.18 . Treat the molecule as two massive points (representing the oxygen atoms) separated by a distance of \(1.21 \times 10^{-10} \mathrm{m}\) . The molar mass of oxygen atoms is \(16.0 \mathrm{g} / \mathrm{mol} .\) (c) Find the rms angular velocity of rotation of an oxygen molecule about either the \(y\) - or \(z\) -axis shown in Fig. 18.15. How does your answer compare to the angular velocity of a typical piece of rapidly rotating machinery \((10,000 \mathrm{rev} / \mathrm{min}) ?\)

In the ideal-gas equation, the number of moles per volume \(n / V\) is simply equal to \(p / R T\) . In the van der Waals equation, solving for \(n / V\) in terms of the pressure \(p\) and temperature \(T\) is somewhat more involved. (a) Show the van der Waals equation can be written as $$\frac{n}{V}=\left(\frac{p+a n^{2} / V^{2}}{R T}\right)\left(1-\frac{b n}{V}\right)$$ (b) The van der Waals parameters for hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{S}\right)\) are \(a=0.448 \mathrm{J} \cdot \mathrm{m}^{3} / \mathrm{mol}^{2}\) and \(b=4.29 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol}\) . Determine the number of moles per volume of \(\mathrm{H}_{2} \mathrm{S}\) gas at \(127^{\circ} \mathrm{C}\) and an absolute pressure of \(9.80 \times 10^{5} \mathrm{Pa}\) as follows: (i) Calculate a first approximation using the ideal-gas equation, \(n / V=p / R T\) . (ii) Substitute this approximation for \(n / V\) into the right-hand side of the equation in part (a). The result is a new, improved approximation for \(n / V\) . (iii) Substitute the new approximation for \(n / V\) into the right-hand side of the equation in (a). The result is a further improved approximation for \(n / V\) . (iv) Repeat step (iii) until successive approximations agree to the desired level of accuracy (in this case, to three significant figures). (c) Compare your final result in part (b) to the result \(p / R T\) obtained using the ideal-gas equation. Which result gives a larger value of \(n / V ?\)

The surface of the sun has a temperature of about 5800 \(\mathrm{K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is 1.67 \(\times 10^{-27} \mathrm{kg} . )\) (b) The escape speed for a particle to leave the gravitational influence of the sun is given by \((2 G M / R)^{1 / 2}\) , where \(M\) is the sun's mass, \(R\) its radius, and \(G\) the gravitational constant (see Example 12.5 of Section \(12.3 ) .\) Use the data in Appendix \(\mathrm{F}\) to calculate this escape speed. (c) Can appreciable quantities of hydrogen escape from the sun? Can any hydrogen escape? Explain.

A balloon whose volume is 750 \(\mathrm{m}^{3}\) is to be filled with hydrogen at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right) .\) (a) If the hydrogen is stored in cylinders with volumes of 1.90 \(\mathrm{m}^{3}\) at a gauge pressure of \(1.20 \times 10^{6} \mathrm{Pa},\) how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if the gas in the balloon and the surrounding air are both at \(15.0^{\circ} \mathrm{C}\) ? The molar mass of hydro\(\operatorname{gen}\left(\mathrm{H}_{2}\right)\) is 2.02 \(\mathrm{g} / \mathrm{mol}\) . The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.23 \(\mathrm{kg} / \mathrm{m}^{3}\) . See Chapter 14 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass 4.00 \(\mathrm{g} / \mathrm{mol}\) ) instead of hydrogen, again at \(15.0^{\circ} \mathrm{C} ?\)

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