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What mass of helium is needed to pressurize an evacuated \(25.0-\mathrm{L}\) tank to a pressure of \(120 .\) atm at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Approximately 488.16 grams of helium is needed to pressurize the evacuated 25.0-L tank to a pressure of 120 atm at \(25^{\circ}\mathrm{C}\).

Step by step solution

01

Convert temperature to Kelvin

Since the given temperature is in Celsius, we need to convert it to Kelvin by adding 273.15 to the Celsius temperature. Therefore, the temperature in Kelvin is: T = 25 + 273.15 = 298.15 K
02

Calculate the number of moles of helium using the ideal gas law

The ideal gas law equation is: PV = nRT We are given P, V, and T, and we need to find n, the number of moles. We will also use R = 0.08206 L atm/mol K, the gas constant for these units. Rearrange the ideal gas law equation to solve for n: n = PV / (RT) Now, substitute the given values for P, V, and T, along with the gas constant R: n = (120 atm)(25.0 L) / (0.08206 L atm/mol K)(298.15 K)
03

Calculate the number of moles of helium

Now, solve the equation to find the number of moles of helium: n = (3000) / (24.6106) ≈ 121.92 moles
04

Calculate the mass of helium

Use the molar mass of helium, which is 4.0026 g/mol, to convert moles to mass: mass = n × molar mass mass = (121.92 moles) × (4.0026 g/mol)
05

Determine the mass of helium needed

Calculate the mass of helium needed to pressurize the tank: mass = 121.92 × 4.0026 ≈ 488.16 g Therefore, approximately 488.16 grams of helium is needed to pressurize the evacuated 25.0-L tank to a pressure of 120 atm at 25°C.

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Key Concepts

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

Gas Pressure Calculations
Understanding gas pressure calculations is essential when working with the behavior of gases, such as in the given exercise which requires us to pressurize a container with helium.

At a molecular level, gas pressure results from the collisions of gas particles with the walls of their container. The more particles or the faster they move (which increases at higher temperatures), the more frequent and forceful their collisions with the container walls, leading to higher pressure.

For practical calculations, the ideal gas law provides a handy tool. The law is usually stated as \( PV = nRT \), connecting the pressure (P), volume (V), and temperature (T) of a gas to the number of moles (n) present and the ideal gas constant (R). By rearranging the formula and solving for the desired variable, one can determine quantities like the volume of a gas necessary to achieve a certain pressure at a given temperature or, as seen in the exercise, the mass of gas needed to pressurize a container.
Molar Mass of Helium
The molar mass of an element like helium is the weight of one mole (6.022 x 1023 atoms) of that substance. Helium, with the symbol He, is the second lightest element on the periodic table and has an atomic number of 2.

The atomic weight of helium is approximately 4.0026 grams per mole. This value is critical when converting between the mass of the gas and the number of moles during calculations. For instance, as illustrated in the exercise, knowing the molar mass allows you to determine the weight of helium you need to achieve a certain number of moles, which in turn can be used together with the ideal gas law to calculate gas pressure-related properties.
Converting Celsius to Kelvin
Temperature conversion between Celsius and Kelvin is a basic but vital step in gas calculations, especially when applying the ideal gas law.

These two scales are related linearly, with the Kelvin scale being an absolute temperature scale starting at absolute zero, unlike Celsius. There is a 273.15 degree difference between the zero of Celsius and the zero of Kelvin. Therefore, to convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. This is crucial because the ideal gas law requires that temperature be in Kelvin for the calculations to be correct.

In our exercise, the given temperature was 25 degrees Celsius, which we converted to Kelvin to use in the ideal gas law formula. Failure to convert the temperature could lead to incorrect results and an incomplete understanding of how temperature affects gas behavior.

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Most popular questions from this chapter

If \(23.2 \mathrm{g}\) of a given gas occupies a volume of \(93.2 \mathrm{L}\) at a particular temperature and pressure, what mass of the gas occupies a volume of \(10.4 \mathrm{L}\) under the same conditions?

Given each of the following sets of values for three of the gas variables, calculate the unknown quantity. a. \(P=21.2\) atm; \(V=142 \mathrm{mL} ; n=0.432\) mol; \(T=\) \(? \mathrm{K}\) b. \(P=?\) atm; \(V=1.23 \mathrm{mL} ; n=0.000115\) mol; \(T=\) \(293 \mathrm{K}\) c. \(P=755 \mathrm{mm}\) Hg; \(V=? \mathrm{mL} ; n=0.473 \mathrm{mol} ; T=\) \(131^{\circ} \mathrm{C}\)

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