An outdoor storage vessel for hydrogen gas with a volume of \(300 . \mathrm{m}^{3}\) is at \(1.5\) atm and \(10 .{ }^{\circ} \mathrm{C}\) at 2:00 AM. By 2:00 PM, the temperature has risen to \(30 .{ }^{\circ} \mathrm{C}\). What is the new pressure of the hydrogen in the vessel?

Short Answer

Expert verified
The new pressure of the hydrogen in the vessel at 2:00 PM is approximately 1.6 atm.

Step by step solution

01

Convert temperatures to Kelvin

Convert the initial and final temperatures from degrees Celsius to Kelvin by adding 273.15 to each temperature. This is necessary because the gas laws require absolute temperatures. Initial temperature (T1): 10°C + 273.15 = 283.15 K Final temperature (T2): 30°C + 273.15 = 303.15 K
02

Apply Charles’s Law

Since the volume of the gas remains constant, use Charles’s Law, which states that the pressure of a given mass of an ideal gas is directly proportional to its Kelvin temperature at constant volume. The formula is: \( \frac{P1}{T1} = \frac{P2}{T2} \).Solving for the new pressure (P2), we get: \( P2 = \frac{P1 \cdot T2}{T1} \)
03

Insert the known values

Insert the initial pressure (P1), initial temperature (T1), and final temperature (T2) into the equation from Step 2. We have \( P1 = 1.5 \) atm, \( T1 = 283.15 \) K, and \( T2 = 303.15 \) K. So, \( P2 = \frac{1.5 \text{ atm} \cdot 303.15 \text{ K}}{283.15 \text{ K}} \)
04

Calculate the final pressure

Perform the calculation from Step 3 to find the final pressure: \( P2 = \frac{1.5 \times 303.15}{283.15} \) atm. Calculate this to get the value of P2 in atm.

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Key Concepts

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

Understanding Gas Laws
Gas laws are foundational principles in chemistry that describe how gases behave under different conditions of temperature, pressure, and volume. These laws are crucial for predicting the behavior of gases in various scenarios, such as in understanding weather patterns, designing engines, and even in our example of a hydrogen gas storage vessel.

One fundamental aspect of these laws is the direct and inverse relationships shared between the gas variables. As seen in our exercise, Charles's Law focuses specifically on the direct relationship between temperature and pressure of a gas when the volume is held constant. This means if the temperature of a gas increases, the pressure will also increase if the volume is not changed.

Utilizing the appropriate gas law as in our exercise, allows us to reliably calculate changes in the gas's behavior when subjected to different temperatures. Simplifying the process of these calculations and understanding the underlying principles can empower students to solve complex problems effortlessly.
The Role of Absolute Temperature in Gas Laws
Absolute temperature is measured in Kelvin and is essential in the study of gas laws because it provides a standard starting point, unlike Celsius or Fahrenheit scales that have arbitrary zero points. Absolute zero in the Kelvin scale represents the theoretical point at which particles have minimal thermal motion, indicating no kinetic energy.

To apply gas laws correctly, as in our exercise, temperatures must be converted to Kelvin. This avoids negative values and ensures the mathematical relationships in the equations remain consistent. Using Kelvin allows us to appreciate the importance of absolute temperature in evaluating how a gas will respond to thermal changes.

By converting Celsius temperatures to Kelvin, we acknowledge the absolute nature of thermal energy in gas behaviors and provide the precision required for accurate calculations, which is necessary for deriving the correct new pressure in a constant volume scenario like in the provided exercise.
Ideal Gas Behavior and Real-world Applications
The concept of ideal gas behavior simplifies the complex interactions between gas particles by assuming ideal conditions: gases consist of point particles that do not attract or repel each other, collisions between gas particles and with the walls of the container are perfectly elastic, and there are no volume effects of the particles themselves.

In reality, no gas perfectly fits these criteria, but many behave closely enough to the ideal model under certain conditions that the model is extremely useful. In our exercise, assuming the hydrogen gas behaves as an ideal gas allows us to apply Charles’s Law for a reliable estimation of how temperature changes influence pressure.

Understanding this model helps explain and predict gas behavior in a variety of real-world applications, from the workings of household refrigerators to the engineering of high-speed jet engines. Always remember the assumptions behind the ideal gas law when making calculations or predictions about real gases, and account for deviations from ideality under extreme conditions of temperature and pressure.

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Most popular questions from this chapter

A piece of lithium metal was added to a flask of water on a day when the atmospheric pressure was \(757.5\) Torr. The lithium reacted completely with the water to produce \(250.0 \mathrm{~mL}\) of hydrogen gas, collected over the water at \(23^{\circ} \mathrm{C}\), at which temperature the vapor pressure of water is \(21.07\) Torr. (a) What is the partial pressure of hydrogen in the collection flask? (b) What mass of lithium metal reacted?

Which starting condition would produce the larger volume of carbon dioxide by combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) with an excess of oxygen gas to produce carbon dioxide and water: (a) \(2.00 \mathrm{~L}\) of \(\mathrm{CH}_{4}(\mathrm{~g})\); (b) \(2.00 \mathrm{~g}\) of \(\mathrm{CH}_{4}(\mathrm{~g})\) ? Justify your answer. The system is maintained at a temperature of \(75^{\circ} \mathrm{C}\) and \(1.00\) atm.

When Robert Boyle conducted his experiments, he measured pressure in inches of mercury (in \(\mathrm{Hg}\) ). On a day when the atmospheric pressure was \(29.85 \mathrm{inHg}\), he trapped some air in the tip of a J-tube (1) and measured the difference in height of the mercury in the two arms of the tube \((b)\). When \(h=12.0\) inches, the height of the gas in the tip of the tube was \(32.0\) in. Boyle then added additional mercury and the level rose in both arms of the tube so that \(h=30.0\) inches \((2)\). (a) What was the height of the air space (in inches) in the tip of the tube in \((2)\) ? (b) What was the pressure of the gas in the tube in (1) and in (2) in inHg?

The root mean square speed, \(v_{\text {rms }}\), of gas molecules was derived in Section 4.10. Using the Maxwell distribution of speeds, we can also calculate the mean speed and most probable (mp) speed of a collection of molecules. The equations used to calculate these two quantities are \(v_{\text {mean }}=(8 R T / \pi M)^{1 / 2}\) and \(v_{\mathrm{mp}}=(2 R T / M)^{1 / 2}\). These values have a fixed relation to one another. (a) Place these three quantities in order of increasing magnitude. (b) Show that the relative magnitudes are independent of the molar mass of the gas. (c) Using the smallest speed as the reference point, rank them by magnitude and determine the ratios of the larger values to the smallest.

The density of a gaseous compound of phosphorus is \(0.943\) \(\mathrm{g} \cdot \mathrm{L}^{-1}\) at \(420 . \mathrm{K}\) when its pressure is 727 Torr. (a) What is the molar mass of the compound? (b) If the compound remains gaseous, what would be its density at \(1.00 \mathrm{~atm}\) and \(298 \mathrm{~K}\) ?

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