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Chapter 3: Problem 29

Argon is an inert gas used in light bulbs to retard the vaporization of the filament. \(A\) certam light-bulb containing argon at \(1.25 \mathrm{~atm}\) and \(18^{\circ} \mathrm{C}\) is heated to \(85^{\circ} \mathrm{C}\) at constant volume. Calculate its final pressure. (a) \(1.53 \mathrm{~atm}\) (b) \(1.25\) atm (c) \(1.35 \mathrm{~atm}\) (d) \(2 \mathrm{~atm}\)

Short Answer

Expert verified
The final pressure of the argon gas when heated to 85°C at constant volume is approximately 1.53 atm (a).

Step by step solution

01

Understand the Ideal Gas Law

The Ideal Gas Law is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the amount of substance (in moles), R is the gas constant, and T is the absolute temperature in Kelvin. For this problem, since the volume and the amount of gas are constant, the relationship between pressure and temperature can be simplified to P1/T1 = P2/T2.
02

Convert Temperature to Kelvin

Convert both initial and final temperatures from Celsius to Kelvin, by adding 273.15 to each of them. Initial temperature: T1 = 18 + 273.15 = 291.15 K. Final temperature: T2 = 85 + 273.15 = 358.15 K.
03

Set Up the Equation

Use the simplified gas law relation (P1/T1 = P2/T2), and input the given values and temperatures in Kelvin to calculate P2. The initial pressure P1 is 1.25 atm, T1 is 291.15 K, and T2 is 358.15 K.
04

Calculate Final Pressure

Rearrange the equation to solve for P2: P2 = P1 * (T2/T1). Substitute in the known values to get P2 = 1.25 atm * (358.15 K / 291.15 K).
05

Perform the Calculation

Multiply the values to get the final pressure, P2: P2 = 1.25 atm * (358.15 K / 291.15 K) = 1.25 atm * 1.23 = 1.5375 atm.
06

Round the Answer

Round the final pressure to two decimal places, as the options are given to two decimal places: The final pressure P2 ≈ 1.54 atm. Thus, the closest answer provided is (a) 1.53 atm.

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Key Concepts

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

Gas Laws
Understanding gas laws is crucial for solving problems related to gas behavior under various conditions. One fundamental principle is the Ideal Gas Law, which relates pressure (P), volume (V), the number of moles of gas (n), temperature (T), and the gas constant (R) as PV = nRT. When dealing with a constant volume and amount of gas, as in our light bulb problem with argon, the Ideal Gas Law simplifies to the direct relationship between pressure and temperature, expressed as P1/T1 = P2/T2. This equation is based on the premise that if the volume and number of moles remain unchanged, the increase or decrease in temperature will proportionally affect the pressure.
Physical Chemistry
Physical Chemistry is the branch of chemistry that deals with the physical properties and behavior of matter, including the Ideal Gas Law and other gas laws. These laws describe how gases react to changes in temperature, pressure, and volume, and understanding them is essential for experimental design, data interpretation, and industry applications, like maintaining the correct conditions inside a light bulb. The Ideal Gas Law problem from our example embodies the dynamic nature of physical chemistry concepts and how they apply to real-world situations.
JEE Chemistry preparation
Preparing for competitive exams like the Joint Entrance Examination (JEE) in India involves mastering various concepts of chemistry, including the Gas Laws. Questions similar to the argon light bulb scenario commonly appear in the Physical Chemistry section of the JEE Chemistry paper. By familiarizing themselves with the step-by-step approach to solving Ideal Gas Law problems, students can enhance their problem-solving skills and increase their speed and accuracy—a must for excelling in such highly competitive exams.
Pressure-Temperature relationship
The Pressure-Temperature relationship, also known as Gay-Lussac's Law, indicates that for a given mass and constant volume of an ideal gas, the pressure exerted by the gas is directly proportional to its absolute temperature. In the exercise concerning the light bulb filled with argon, we observed this relationship: as the temperature increased from 18°C to 85°C, the pressure within the bulb also rose. This is why converting Celsius to Kelvin becomes a crucial step, ensuring that we compare absolute temperatures, thus allowing a proper application of the law to determine the final gas pressure.

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Most popular questions from this chapter

Under which of the following sets of conditions is a real gas expected to deviate from ideal behaviour? (I) High pressure, small volume (II) High temperature, low pressure (III) Low temperature, high pressure (a) only I (b) only II (c) only III (d) I and III both

\(56 \mathrm{~g}\) of nitrogen and \(96 \mathrm{~g}\) of oxygen are mixed isothermaly and at a total pressure of \(10 \mathrm{~atm} .\) The partial pressures of oxygen and nitrogen (in atm) are respectively : (a) 4,6 (b) 5,5 (c) 2,8 (d) 6,4

"Equal volumes of all gases at the same temperature and pressure contain equal number of particles." This statement is a direct consequence of : (a) Avogadro's law (b) Charle's law (c) Ideal gas equation (d) Law of partial pressure

Two flasks \(A\) and \(B\) have equal volumes. \(A\) is maintained at \(300 \mathrm{~K}\) and \(B\) at \(600 \mathrm{~K}\), while \(A\) contains \(\mathrm{H}_{2}\) gas, \(B\) has an equal mass of \(\mathrm{CO}_{2}\) gas. Find the ratio of total K.E. of gases in flask \(A\) to that of \(B\). (a) \(1: 2\) (b) \(11: 1\) (c) \(33: 2\) (d) \(55: 7\)

The rate of diffusion of a gas is proportional to: (a) \(\frac{P}{\sqrt{d}}\) (b) \(\frac{P}{d}\) (c) \(\sqrt{\frac{P}{d}}\) (d) \(\frac{\sqrt{P}}{d}\)
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