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Chapter 8: Problem 33

Calculate the final pressure, in torr, for each of the following, if \(n\) and \(V\) are constant: a. A gas with an initial pressure of 1500 Torr at \(175^{\circ} \mathrm{C}\) is cooled to \(20^{\circ} \mathrm{C}\). b. A gas in an aerosol can with an initial pressure of 3.10 atm at \(24^{\circ} \mathrm{C}\) is heated to \(45^{\circ} \mathrm{C}\).

Short Answer

Expert verified
a. 981.56 Torr, b. 2520 Torr.

Step by step solution

01

Understand the Ideal Gas Law

Use the Ideal Gas Law in the form of the combined gas law since the number of moles and the volume are constant. The combined gas law is given by: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] where \(P\) is pressure and \(T\) is temperature in Kelvin.
02

Convert temperatures to Kelvin

Convert all temperatures to Kelvin using the formula: \( T(K) = T(掳C) + 273.15 \). For part a: Initial temperature: \( T_1 = 175 + 273.15 = 448.15 \text{ K} \). Final temperature: \( T_2 = 20 + 273.15 = 293.15 \text{ K} \). For part b: Initial temperature: \( T_1 = 24 + 273.15 = 297.15 \text{ K} \). Final temperature: \( T_2 = 45 + 273.15 = 318.15 \text{ K} \).
03

Apply the combined gas law for part a

For part a: \( P_1 = 1500 \text{ Torr} \), \( T_1 = 448.15 \text{ K} \), and \( T_2 = 293.15 \text{ K} \). Use the combined gas law to find \( P_2 \): \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \frac{T_2}{T_1} = 1500 \times \frac{293.15}{448.15} \approx 981.56 \text{ Torr} \]
04

Convert initial pressure for part b

For part b, convert the initial pressure from atm to Torr, knowing that 1 atm = 760 Torr: \( P_1 = 3.10 \times 760 = 2356 \text{ Torr} \).
05

Apply the combined gas law for part b

For part b: \( P_1 = 2356 \text{ Torr} \), \( T_1 = 297.15 \text{ K} \), and \( T_2 = 318.15 \text{ K} \). Use the combined gas law to find \( P_2 \): \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \frac{T_2}{T_1} = 2356 \times \frac{318.15}{297.15} \approx 2520 \text{ Torr} \]

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Key Concepts

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

ideal gas law
The ideal gas law is a fundamental equation in chemistry. It relates the pressure, volume, temperature, and number of moles of a gas. The formula is usually written as \(PV = nRT\). However, when the volume and number of moles are constant, we use the combined gas law: \[\frac{P_1}{T_1} = \frac{P_2}{T_2}\]. This form helps us calculate changes in pressure and temperature while keeping volume and moles constant. Remember: pressure is in units like atm or Torr, and temperature must be in Kelvin for the equations to work correctly.
temperature conversion
Converting temperature from Celsius to Kelvin is crucial in gas law calculations. The formula is simple: \(T(K) = T(掳C) + 273.15\). This adjustment accounts for the zero point in the Kelvin scale, aligned with absolute zero where molecular motion theoretically stops.
For example:
  • Initial temperature: 175掳C = 175 + 273.15 = 448.15 K
  • Final temperature: 20掳C = 20 + 273.15 = 293.15 K
Always implement this step to avoid errors in your calculations.
pressure conversion
Properly converting pressure units can often avoid mistakes in gas law problems. For example, when pressures are given in atmospheres (atm) but the calculation requires Torr:
  • 1 atm = 760 Torr
Doing conversions:
  • Initial pressure: 3.10 atm = 3.10 脳 760 = 2356 Torr
Ensure to convert all your units correctly before substituting values into the combined gas law.

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Most popular questions from this chapter

A gas with a volume of \(4.0 \mathrm{~L}\) is in a closed container. Indicate the changes (increases, decreases, does not change) in its pressure when the volume undergoes the following changes at the same temperature and amount of gas: a. The volume is compressed to \(2.0 \mathrm{~L}\). b. The volume expands to \(12 \mathrm{~L}\) c. The volume is compressed to \(0.40 \mathrm{~L}\).

Use the words inspiration and expiration to describe the part of the breathing cycle that occurs as a result of each of the following: a. The diaphragm contracts. b. The volume of the lungs decreases. c. The pressure within the lungs is less than that of the atmosphere.

A sample of oxygen \(\left(\mathrm{O}_{2}\right)\) has a volume of \(30.0 \mathrm{~L}\) at a pressure of \(760 . \mathrm{mmHg}\). What is the final volume, in liters, of the gas at each of the following pressures, if there is no change in temperature and amount of gas? a. \(625 \mathrm{mmHg}\) b. 3.0 atm c. 0.800 atm d. 350 Torr

Calculate the final pressure, in atmospheres, for each of the following, if \(V\) and \(n\) do not change: a. A gas with an initial pressure of \(1.20 \mathrm{~atm}\) at \(75^{\circ} \mathrm{C}\) is cooled to \(-32^{\circ} \mathrm{C}\) b. A sample of \(\mathrm{N}_{2}\) with an initial pressure of \(780 . \mathrm{mmHg}\) at \(-75^{\circ} \mathrm{C}\) is heated to \(28^{\circ} \mathrm{C}\)

Indicate whether the final volume of gas in each of the following is the same, larger, or smaller than the initial volume, if pressure and amount of gas do not change: a. A volume of \(505 \mathrm{~mL}\) of air on a cold winter day at \(-15^{\circ} \mathrm{C}\) is breathed into the lungs, where body temperature is \(37^{\circ} \mathrm{C}\). b. The heater used to heat the air in a hot-air balloon is turned off. c. A balloon filled with helium at the amusement park is left in a car on a hot day.
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