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Chapter 15: Problem 3

\(\bullet\) A 3.00 L tank contains air at 3.00 atm and \(20.0^{\circ} \mathrm{C}\) . The tank is sealed and cooled until the pressure is 1.00 atm. (a) What is the temperature then in degrees Celsius, assuming that the volume of the tank is constant? (b) If the temperature is kept at the value found in part (a) and the gas is compressed, what is the volume when the pressure again becomes 3.00 atm?

Short Answer

Expert verified
(a) -175.43 掳C; (b) 1.00 L

Step by step solution

01

Understanding the Problem

We are given a tank with an initial pressure and temperature. The goal is to find the final temperature when the pressure decreases, assuming constant volume, and then to find the volume when the pressure is returned to its initial value at the new temperature.
02

Using Ideal Gas Law for Part (a)

Using the ideal gas law, remember the relation for constant volume: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). We know \( P_1 = 3.00 \ \text{atm}, T_1 = 20.0\ ^\circ \text{C} = 293.15 \ \text{K} \), and \( P_2 = 1.00 \ \text{atm} \).
03

Rearranging for T2

Rearrange the formula to solve for \( T_2 \): \( T_2 = \frac{P_2 \cdot T_1}{P_1} \). Plug in the known values: \( T_2 = \frac{1.00 \ \text{atm} \cdot 293.15 \ \text{K}}{3.00 \ \text{atm}} \).
04

Calculating T2

Compute \( T_2 \): \( T_2 = 97.717 \ \text{K} \). Convert the temperature to Celsius: \( T_2 = 97.717 \ \text{K} - 273.15 = -175.43 \ ^\circ \text{C} \).
05

Using Ideal Gas Law for Part (b)

For compression at constant temperature \( T = -175.43 \ ^\circ \text{C} = 97.717 \ \text{K} \), we use \( \frac{P_1 V_1}{T} = \frac{P_2 V_2}{T} \). Simplify as \( P_1 V_1 = P_2 V_2 \) since the temperature is constant.
06

Rearranging for V2

Solve for \( V_2 \): \( V_2 = \frac{P_1 V_1}{P_2} \). Substitute \( P_1 = 1.00 \ \text{atm}, V_1 = 3.00 \ \text{L}, \) and \( P_2 = 3.00 \ \text{atm} \).
07

Calculating V2

Calculate the volume: \( V_2 = \frac{1.00 \ \text{atm} \cdot 3.00 \ \text{L}}{3.00 \ \text{atm}} = 1.00 \ \text{L} \).

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Key Concepts

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

Pressure and Temperature Relationship
In the realm of gases, there's a fundamental rule connecting pressure and temperature, known as Gay-Lussac's Law. This principle tells us that if you have a constant volume, like a sealed tank, the pressure of a gas is directly proportional to its temperature, measured in Kelvin.
This means that if you increase the temperature of the gas, its pressure will also rise, provided the volume doesn't change. Conversely, cooling the gas leads to a drop in pressure.
  • This relationship is mathematically represented as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), where \(P\) is pressure and \(T\) is temperature in Kelvin.
  • It's essential to convert temperatures from Celsius to Kelvin by adding 273.15 because Kelvin temperatures prevent negative values that would create complications in calculations.
  • Understanding this relationship helps us predict how gases will behave under different thermal conditions while within a closed environment.
Constant Volume Process
The idea of a constant volume process is crucial in understanding gas behavior when no change in space occurs. This means the container the gas is in doesn't expand or contract; in our problem, it's a sealed tank.
  • In such scenarios, when a gas is heated, its pressure rises without any increase in volume, due to the particles moving faster and hitting the walls more often.
  • Conversely, if cooled, the pressure decreases as particles slow down.
  • In the initial part of our exercise, the volume remained fixed, and hence, using the pressure and temperature relationship, we could determine the new temperature when pressure changed.

This concept makes it easier to understand real-life applications like the working principle of pressure cookers or car engines.
Gas Compression
Compressing a gas involves decreasing its volume while increasing its pressure. Here, we're looking at gas compression after it has cooled.
According to the ideal gas law, if the temperature remains constant, which was the case in part "b" of our exercise, the product of pressure and volume is constant.
  • This relationship is expressed as \( P_1 V_1 = P_2 V_2 \), signaling that when pressure increases, volume must decrease to maintain the equation's balance.
  • This principle helps calculate the new volume when a gas is compressed from a lower pressure back to its initial, higher pressure.
  • Practical examples include air compressors and gas cylinders, which rely on reducing volume to store gases at high pressure.
Temperature Conversion
Temperature conversion is essential in scientific calculations involving gases because physical laws require temperature in Kelvin.
The Kelvin scale starts at absolute zero, the coldest possible temperature, making it a more natural fit for gas law calculations.
  • To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. This adjustment shifts the zero points of the scales relative to each other, bridging the gap between everyday and scientific use.
  • Having temperatures in Kelvin ensures that all values in equations are positive, avoiding errors in physical interpretation and mathematical calculations.
  • It's a step often overlooked, but critical for accuracy, especially in our exercise where initial values were provided in degrees Celsius.

Thus, always do the conversion first before plugging temperatures into any gas law equation for reliable results.

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Most popular questions from this chapter

\(\bullet$$\bullet\) The effect of altitude on the lungs. (a) Calculate the change in air pressure you will experience if you climb a 1000 \(\mathrm{m}\) moun- tain, assuming that the temperature and air density do not change over this distance and that they were \(22^{\circ} \mathrm{Cand} 1.2 \mathrm{kg} / \mathrm{m}^{3},\) respectively, at the bottom of the mountain. (b) If you took a 0.50 \(\mathrm{L}\) breath at the foot of the mountain and managed to hold it until you reached the top, what would be the volume of this breath when you exhaled it there?

\(\bullet\) You are keeping 1.75 moles of an ideal gas in a container surrounded by a large ice-water bath that maintains the temperature of the gas at \(0.00^{\circ} \mathrm{C}\) (a) How many joules of work would have to be done on this gas to compress its volume from 4.20 \(\mathrm{L}\) to 1.35 \(\mathrm{L} ?\) (b) How much heat came into (or out of) the gas during this process? Was it into or out of?

\(\bullet$$\bullet\) At an altitude of \(11,000 \mathrm{m}\) (a typical cruising altitude for a jet airliner), the air temperature is \(-56.5^{\circ} \mathrm{C}\) and the air density is 0.364 \(\mathrm{kg} / \mathrm{m}^{3} .\) What is the pressure of the atmosphere at that altitude? The molar mass of air is 28.8 \(\mathrm{g} / \mathrm{mol}\) .

\(\bullet\) The gas inside a balloon will always have a pressure nearly equal to atmospheric pressure, since that is the pressure applied to the outside of the balloon. You fill a balloon with helium (a nearly ideal gas) to a volume of 0.600 L at a temperature of \(19.0^{\circ} \mathrm{C}\) . What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen \((77.3 \mathrm{K})\) ?

\(\bullet$$\bullet\) The atmosphere of the planet Mars is 95.3\(\%\) carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and about 0.03\(\%\) water vapor. The atmospheric pressure is only about 600 \(\mathrm{Pa}\) , and the surface temperature varies from \(-30^{\circ} \mathrm{C}\) to \(-100^{\circ} \mathrm{C}\) . The polar ice caps contain both \(\mathrm{CO}_{2}\) ice and water ice. Could there be liquid \(\mathrm{CO}_{2}\) on the surface of Mars? Could there be liquid water? Why or why not?
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