/*! 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 29 Argon is an inert gas used in li... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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掳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 鈮� 1.54 atm. Thus, the closest answer provided is (a) 1.53 atm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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鈥攁 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掳C to 85掳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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A balloon weighing \(50 \mathrm{~kg}\) is filled with \(685 \mathrm{~kg}\) of helium at 1 atm pressure and \(25^{\circ} \mathrm{C}\). What will be its pay load if it displaced \(5108 \mathrm{~kg}\) of air? (a) \(4373 \mathrm{~kg}\) (b) \(4423 \mathrm{~kg}\) (c) \(5793 \mathrm{~kg}\) (d) none of these

The total pressure of a mixture of oxygen and hydrogen is \(1.0 \mathrm{~atm} .\) The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of \(0.40 \mathrm{~atm}\) when measured at the same values of \(T\) and \(V\) as the original mixture. What was the composition of the original mixture in mole per cent? (a) \(x_{\mathrm{O}_{2}}=0.2 ; x_{\mathrm{H}_{2}}=0.8\) (b) \(x_{\mathrm{O}_{2}}=0.4 ; x_{\mathrm{H}_{2}}=0.6\) (c) \(x_{\mathrm{O}_{2}}=0.6 ; x_{\mathrm{H}_{2}}=0.4\) (d) \(x_{\mathrm{O}_{2}}=0.8 ; x_{\mathrm{H}_{2}}=0.2\)

At constant volume, for a fixed number of moles of a gas, the pressure of the gas increases with increase in temperature due to: (a) Increase in the average molecular speed (b) Increase rate of collision amongst molecules (c) Increase in molecular attraction (d) Decrease in mean free path

At \(273 \mathrm{~K}\) temp. and 9 atm pressure, the compressibility for a gas is \(0.9 .\) The volume of 1 milli-moles of gas at this temperature and pressure is : (a) \(2.24\) litre (b) \(0.020 \mathrm{~mL}\) (c) \(2.24 \mathrm{~mL}\) (d) \(22.4 \mathrm{~mL}\)

Three flasks of equal volumes contain \(\mathrm{CH}_{4}, \mathrm{CO}_{2}\) and \(\mathrm{Cl}_{2}\) gases respectively. They will contain equal number of molecules if : (a) the mass of all the gases is same (b) the moles of all the gas is same but temperature is different (c) temperature and pressure of all the flasks are same (d) temperature, pressure and masses same in the flasks
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.