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Suppose you seal a rigid container that has been open to air at sea level when the temperature is \(0^{\circ} \mathrm{C}(273 \mathrm{K}) .\) The pressure inside the sealed container is now exactly equal to the outside air pressure: \(10^{5} \mathrm{N} / \mathrm{m}^{2}\). a. What would be the pressure inside the container if it were left sitting in the desert shade where the surrounding air temperature was \(50^{\circ} \mathrm{C}(323 \mathrm{K}) ?\) b. What would be the pressure inside the container if it were left sitting out in an Antarctic night where the surrounding air temperature was \(-70^{\circ} \mathrm{C}(203 \mathrm{K}) ?\) c. What would you observe in each case if the walls of the container were not rigid?

Short Answer

Expert verified
a. Pressure in the desert is \( 1.1836 \times 10^5 \, N/m^2 \). b. Pressure in Antarctica is \( 7.4385 \times 10^4 \, N/m^2 \). c. If non-rigid, the container would expand in desert and contract in Antarctica.

Step by step solution

01

Understand the Ideal Gas Law

The problem can be approached using the Ideal Gas Law: \[ PV = nRT \]where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
02

Set Up Initial Conditions

Initially, the temperature inside the container is \( 273 \, K \) and the pressure is \( 10^5 \, N/m^2 \). This gives us: \[ P_1 = 10^5 \, N/m^2 \] and \[ T_1 = 273 \, K \]
03

Apply the Ideal Gas Law for the Desert Case (323 K)

Use the relationship derived from the Ideal Gas Law:\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]Given: \[ P_1 = 10^5 \, N/m^2 \]\[ T_1 = 273 \, K \]\[ T_2 = 323 \, K \] We need to find \( P_2 \):\[ \frac{10^5 \, N/m^2}{273 \, K} = \frac{P_2}{323 \, K} \] Solving for \( P_2 \):\[ P_2 = 10^5 \, N/m^2 \times \frac{323 \, K}{273 \, K} \]\[ P_2 \approx 1.1836 \times 10^5 \, N/m^2 \]
04

Apply the Ideal Gas Law for the Antarctic Case (203 K)

Given: \[ P_1 = 10^5 \, N/m^2 \]\[ T_1 = 273 \, K \]\[ T_2 = 203 \, K \] We need to find \( P_3 \):\[ \frac{10^5 \, N/m^2}{273 \, K} = \frac{P_3}{203 \, K} \] Solving for \( P_3 \):\[ P_3 = 10^5 \, N/m^2 \times \frac{203 \, K}{273 \, K} \]\[ P_3 \approx 7.4385 \times 10^4 \, N/m^2 \]
05

Observe the Non-Rigid Container Case

If the container walls were not rigid, the volume would change to balance the internal and external pressures. In the desert, where the temperature is higher, the gas inside would expand, potentially causing the container to swell. In the Antarctic night, where the temperature is lower, the gas inside would contract, potentially causing the container to shrink.

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Key Concepts

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

Pressure-Temperature Relationship
The pressure and temperature of a gas are closely related, which is a concept crucial to understanding many thermodynamic processes. According to the Ideal Gas Law, which is given by the formula \[ PV = nRT \]where:
  • \( P \) is the pressure
  • \( V \) is the volume
  • \( n \) is the number of moles of gas
  • \( R \) is the universal gas constant
  • \( T \) is the temperature measured in Kelvin

When dealing with problems where the container's volume (\( V \)) remains constant (such as in a rigid container), the relationship between pressure and temperature can be simplified to \[ \frac{P1 }{ T1 }= \frac{P2 }{ T2 } \]This implies that for a given mass of gas at constant volume, the pressure is directly proportional to its absolute temperature. Therefore, as the temperature increases, the pressure also increases, and vice versa.
In the provided exercise, we utilized this direct proportional relationship to determine the pressures in different environments: in the desert and in Antarctic conditions.
Kelvin Temperature Scale
To correctly apply the Ideal Gas Law, we must use the Kelvin scale for temperature measurements. The Kelvin scale starts at absolute zero, which is the theoretical point where particles have minimal thermal motion. The conversion between Celsius and Kelvin is straightforward:
  • \[ K = °C + 273.15 \]
In the exercise, the given temperatures were already converted to the Kelvin scale, making it easy to use them in our calculations. Using temperatures in Kelvin is essential because it provides a true zero point that aligns with the behavior we see in gases as they cool down (decreasing in pressure) and heat up (increasing in pressure).
For example, when the ambient temperature in the exercise was \(0° C (273 K)\), if the temperature in the desert rises to \( 50° C (323 K) \), using the Kelvin temperature scale makes it clear how to calculate the new pressures using the Ideal Gas Law.
Thermal Expansion and Contraction
Thermal expansion and contraction describe how substances change in size with temperature variations. For gases, this behavior is explained well by the Ideal Gas Law. When the temperature increases, gas molecules move faster and tend to spread out more, which is expansion. Conversely, when the temperature drops, gas molecules slow down and occupy less space, resulting in contraction.
In the context of the non-rigid container in the exercise, if the walls of the container could move, here's what would happen:
  • In the desert scenario (higher temperature), the gas would expand due to increased molecular activity, potentially causing the container to bulge outwards.
  • In the Antarctic night scenario (lower temperature), the gas would contract, and the container might shrink or collapse inward.
Understanding these principles helps to explain observations in everyday life, such as why car tires can look flattened on a cold morning and reinflate as the day warms up.

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