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Chapter 8: Problem 39

A tank contains isoflurane, an inhaled anesthetic, at a pressure of 1.8 atm and \(5^{\circ} \mathrm{C}\). What is the pressure, in atmospheres, if the gas is warmed to a temperature of \(22^{\circ} \mathrm{C},\) if \(V\) and \(n\) do not change?

Short Answer

Expert verified
The final pressure is approximately 1.91 atm.

Step by step solution

01

- Identify the given variables

The initial pressure, temperature, final temperature are given. Let initial pressure = 1.8 atm, initial temperature \(T_1 = 5 \degree C\), final temperature \(T_2 = 22 \degree C\).
02

- Convert temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin using the formula \(K = \degree C + 273.15\). Thus, \(T_1 = 5 + 273.15 = 278.15\) K and \(T_2 = 22 + 273.15 = 295.15\) K.
03

- Use the ideal gas law formula

Since volume and number of moles are constant, we can use the combined gas law: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] Rearrange to find the final pressure \(P_2\): \[ P_2 = P_1 \frac{T_2}{T_1} \]
04

- Substitute the known values

Substitute the known values into the equation: \[ P_2 = 1.8 \text{ atm} \frac{295.15 \text{ K}}{278.15 \text{ K}} \]
05

- Calculate the final pressure

Perform the calculation to find \(P_2\): \[ P_2 = 1.8 \text{ atm} \times 1.061 \approx 1.91 \text{ atm} \]

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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 and physics that relates the pressure, volume, temperature, and number of moles of a gas. It is represented by the formula \( PV = nRT \). In this formula, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of the gas, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. The ideal gas law helps predict the behavior of a gas under different conditions, assuming the gas behaves ideally. An ideal gas perfectly follows the assumptions of the kinetic molecular theory and does not account for intermolecular forces. This law is particularly useful in calculations involving gases where the pressure, volume, or temperature changes, but the number of moles remains constant.
Pressure and Temperature Relationship
The relationship between pressure and temperature of a gas at constant volume is given by Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature (in Kelvin). This can be expressed as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). This means if you increase the temperature of a gas, its pressure will also increase if the volume does not change. This relationship is part of the Combined Gas Law, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law. To solve for the new pressure when temperature changes:
  • First, convert temperatures from Celsius to Kelvin.
  • Then use the equation \( P_2 = P_1 \frac{T_2}{T_1} \).
  • Substitute the known values to find the new pressure.
Kelvin Temperature Scale
The Kelvin temperature scale is the absolute thermodynamic temperature scale used in scientific measurements. It starts at absolute zero, the theoretically lowest possible temperature where molecular motion stops. To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature. For example, \( 5 \text{ 掳C} = 5 + 273.15 = 278.15 \text{ K} \) and \( 22 \text{ 掳C} = 22 + 273.15 = 295.15 \text{ K} \). The Kelvin scale is crucial in gas law calculations because it ensures temperature measurements start from an absolute reference point, avoiding negative temperatures and providing consistency in equations involving temperature. Always remember: never use degrees when referring to Kelvin, simply say 'Kelvin' (e.g., 300 K, not 300 掳K).

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Most popular questions from this chapter

A sample containing \(4.80 \mathrm{~g}\) of \(\mathrm{O}_{2}\) gas has an initial volume of \(15.0 \mathrm{~L}\). What is the final volume, in liters, when each of the following occurs and pressure and temperature do not change? a. A sample of 0.500 mole of \(\mathrm{O}_{2}\) is added to the \(4.80 \mathrm{~g}\) of \(\mathrm{O}_{2}\) in the container. b. A sample of \(2.00 \mathrm{~g}\) of \(\mathrm{O}_{2}\) is removed. c. A sample of \(4.00 \mathrm{~g}\) of \(\mathrm{O}_{2}\) is added to the \(4.80 \mathrm{~g}\) of \(\mathrm{O}_{2}\) gas in the container.

A sample containing 1.50 moles of Ne gas has an initial volume of \(8.00 \mathrm{~L}\). What is the final volume, in liters, when each of the following occurs and pressure and temperature do not change? a. A leak allows one-half of Ne atoms to escape. b. A sample of 3.50 moles of Ne is added to the 1.50 moles of Ne gas in the container. c. A sample of \(25.0 \mathrm{~g}\) of Ne is added to the 1.50 moles of Ne gas in the container.

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 \(0.40 \mathrm{~L}\).

A sample of Ar gas has a volume of \(5.40 \mathrm{~L}\) with an unknown pressure. The gas has a volume of \(9.73 \mathrm{~L}\) when the pressure is 3.62 atm, with no change in temperature or amount of gas. What was the initial pressure, in atmospheres, of the gas?

Bacteria and viruses are inactivated by temperatures above \(135^{\circ} \mathrm{C}\). An autoclave contains steam at \(1.00 \mathrm{~atm}\) and \(100^{\circ} \mathrm{C}\). What is the pressure, in atmospheres, when the temperature of the steam in the autoclave reaches \(135^{\circ} \mathrm{C},\) if \(V\) and \(n\) do not change?
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