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Chapter 5: Problem 60

A compressed gas cylinder contains \(1.00 \times 10^{3} \mathrm{~g}\) argon gas. The pressure inside the cylinder is \(2050 .\) psi (pounds per square inch) at a temperature of \(18^{\circ} \mathrm{C}\). How much gas remains in the cylinder if the pressure is decreased to \(650 .\) psi at a temperature of \(26^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The mass of argon gas remaining in the cylinder when the pressure is decreased to 650 psi at a temperature of 26 °C is approximately 309.4 g.

Step by step solution

01

Write down the given information and Ideal Gas Law

We are given the following: Initial Information: - m1 = 1.00 x 10^3 g (mass of argon gas initially) - P1 = 2050 psi (initial pressure) - T1 = 18°C (initial temperature) Final Information: - P2 = 650 psi (final pressure) - T2 = 26°C (final temperature) The Ideal Gas Law is given by: \(PV = nRT\)
02

Convert the given information into the appropriate units

We first need to convert all the given information into appropriate units before using the Ideal Gas Law. 1. Convert temperature from Celsius to Kelvin: T1(K) = T1(°C) + 273.15 = 18 + 273.15 = 291.15 K T2(K) = T2(°C) + 273.15 = 26 + 273.15 = 299.15 K 2. Convert pressure from psi to atm, using the conversion factor 1 atm = 14.696 psi: P1(atm) = 2050 psi / 14.696 = 139.53 atm P2(atm) = 650 psi / 14.696 = 44.26 atm 3. Calculate the initial gas moles using the molar mass of argon (39.95 g/mol): n1 = m1 / M_m = (1.00 x 10^3 g) / 39.95 g/mol = 25.03 mol
03

Use the Ideal Gas Law to relate initial and final states

Since the volume of the cylinder doesn't change, we can set the initial and final states equal to each other. \(\frac{P_1V}{n_1T_1} = \frac{P_2V}{n_2T_2}\) Where n2 is the amount of gas remaining in the cylinder at the final conditions (in moles). We want to solve for n2, so we rearrange the equation: \(n_2 = n_1\frac{P_2T_1}{P_1T_2}\) Now we can plug in the values we found in step 2.
04

Calculate the amount of gas remaining in the cylinder

Using the equation from step 3, plug in the given values: \(n_2 = 25.03\frac{44.26 \times 291.15}{139.53 \times 299.15}\) \(n_2 \approx 7.746 \mathrm{mol}\) We found that the amount of gas remaining in the cylinder is approximately 7.746 moles.
05

Calculate the mass of gas remaining in the cylinder

To find the mass of the gas remaining in the cylinder, we can use the molar mass of argon: m2 = n2 × M_m m2 = 7.746 mol × 39.95 g/mol ≈ 309.4 g So, approximately 309.4 g of argon gas remains in the cylinder when the pressure is decreased to 650 psi at a temperature of 26 °C.

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Key Concepts

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

Argon Gas Properties
Argon is a noble gas, which means it is inert and does not readily react with other elements. Its atomic number is 18, and it's used in applications ranging from lighting to welding due to its non-reactive nature. When discussing argon in the context of the Ideal Gas Law, it is important to know that argon is monatomic and has a molar mass of approximately 39.95 g/mol. As with all gases, the behavior of argon under varying pressures and temperatures can be predicted by gas laws.

As shown in the exercise, when working with argon or any gas, we often use the gas properties alongside the Ideal Gas Law to resolve practical problems such as calculating the remaining quantity of gas under different conditions. To do this effectively, It is imperative to consider the standard conditions of temperature and pressure, which are 273.15 K (0°C) and 1 atm (101.325 kPa) respectively, for any calculations involving gas quantities.
Gas Laws
Gas laws are mathematical relationships between the pressure, volume, temperature, and number of moles of a gas that describe how gases behave under various conditions. The Ideal Gas Law, expressed as PV = nRT, combines several individual gas laws and is fundamental for predicting the behavior of an ideal gas. In this equation, P represents pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is temperature.

When analyzing problems like the one provided, we use the Ideal Gas Law to relate the initial and final states of a gas under constant volume. We can also infer from the Ideal Gas Law that at a fixed volume, the pressure of a gas is directly proportional to its temperature, assuming the amount of gas remains the same. This is known as Gay-Lussac's Law, a subset of the broader gas laws framework.
Pressure and Temperature Relationship
The relationship between pressure and temperature of a gas is described by Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature when the volume is constant. This implies that as the temperature increases, so does the pressure, and vice versa, as long as the amount of gas doesn't change.

Our exercise demonstrates this concept. We have a compressed cylinder of argon gas where the pressure changes as a result of a temperature alteration. The exercise guides us through using the Ideal Gas Law to calculate how the amount of gas changes with altering pressure and temperature. To find the new quantity of gas once conditions change, we use the initial and final states of pressure and temperature. The computation confirms that under decreased pressure and slightly increased temperature, the amount of gas, as measured in moles, decreases – reflecting the principles of the pressure-temperature relationship in gases.

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Most popular questions from this chapter

The total mass that can be lifted by a balloon is given by the difference between the mass of air displaced by the balloon and the mass of the gas inside the balloon. Consider a hot-air balloon that approximates a sphere \(5.00 \mathrm{~m}\) in diameter and contains air heated to \(65^{\circ} \mathrm{C}\). The surrounding air temperature is \(21^{\circ} \mathrm{C}\). The pressure in the balloon is equal to the atmospheric pressure, which is 745 torr. a. What total mass can the balloon lift? Assume that the average molar mass of air is \(29.0 \mathrm{~g} / \mathrm{mol}\). (Hint: Heated air is less dense than cool air.) b. If the balloon is filled with enough helium at \(21^{\circ} \mathrm{C}\) and 745 torr to achieve the same volume as in part a, what total mass can the balloon lift? c. What mass could the hot-air balloon in part a lift if it were on the ground in Denver, Colorado, where a typical atmospheric pressure is 630 . torr?

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Silane, \(\mathrm{SiH}_{4}\), is the silicon analogue of methane, \(\mathrm{CH}_{4}\). It is prepared industrially according to the following equations: $$ \begin{aligned} \mathrm{Si}(s)+3 \mathrm{HCl}(g) & \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g) \\ 4 \mathrm{HSiCl}_{3}(l) & \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l) \end{aligned} $$ a. If \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}(d=1.34 \mathrm{~g} / \mathrm{mL})\) is isolated when \(15.0 \mathrm{~L}\) \(\mathrm{HCl}\) at \(10.0 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3} ?\) b. When \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at \(10.0\) atm and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is \(93.1 \%\) ?

The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot \(\mathrm{CuO}(s)\) : $$ \text { Compound } \underset{\text { Cvoss }}{\stackrel{\text { Hot }}{\longrightarrow}} \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ The product gas is then passed through a concentrated solution of \(\mathrm{KOH}\) to remove the \(\mathrm{CO}_{2}\). After passage through the \(\mathrm{KOH}\) solution, the gas contains \(\mathrm{N}_{2}\) and is saturated with water vapor. In a given experiment a \(0.253-g\) sample of a compound produced \(31.8 \mathrm{~mL} \mathrm{~N}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\) and 726 torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8\) torr.)

A gauge on a compressed gas cylinder reads 2200 psi (pounds per square inch; 1 atm \(=14.7\) psi). Express this pressure in each of the following units. a. standard atmospheres b. megapascals (MPa) c. torr
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