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A 20.0-L tank contains \(4.86 \times 10{^-}{^4}\) kg of helium at 18.0\(^\circ\)C. The molar mass of helium is 4.00 g/mol. (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

Short Answer

Expert verified
(a) 0.1215 moles; (b) 1477.3 Pa, 0.0146 atm.

Step by step solution

01

Convert mass to moles

First, let's convert the mass of helium gas into moles using the molar mass. The formula to find the number of moles (\( n \)) is \( n = \frac{m}{M} \), where \( m \) is the mass and \( M \) is the molar mass. Here, \( m = 4.86 \times 10^{-4} \) kg = 0.486 g and \( M = 4.00 \) g/mol. So, \( n = \frac{0.486}{4.00} \approx 0.1215 \) moles.
02

Calculate pressure using the Ideal Gas Law

To find the pressure, use the Ideal Gas Law \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume in liters, \( n \) is the number of moles, \( R \) is the ideal gas constant (8.314 J/(mol·K)), and \( T \) is the temperature in Kelvin. Convert 18.0°C to Kelvin (\( T = 18 + 273.15 = 291.15 \) K). Using \( n = 0.1215 \) moles, \( V = 20.0 \) L = 0.020 m³, and \( R = 8.314 \): \[ P = \frac{nRT}{V} = \frac{0.1215 \times 8.314 \times 291.15}{0.020} = 1477.3 \text{ Pa} \]
03

Convert pressure to atmospheres

Now, convert the pressure from pascals to atmospheres. The conversion is \( 1 \text{ atm} = 101325 \text{ Pa} \). So, \( P = \frac{1477.3}{101325} \approx 0.0146 \text{ atm} \).

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Key Concepts

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

Moles Calculation
Understanding how to calculate the number of moles in a substance is essential when working with gases, like helium. Moles represent a unit for counting atoms, much like a dozen represents twelve. To determine the number of moles ( n ), you use the formula: \( n = \frac{m}{M} \). Here, m is the mass of the substance, and M is the molar mass of the substance.

In the given exercise, the mass of helium is provided as \( 4.86 \times 10^{-4} \ \text{kg} \) which can be converted to grams, resulting in 0.486 g. With helium's molar mass being 4.00 g/mol, you can plug these values into the formula:
  • Mass ( m ) = 0.486 g
  • Molar mass ( M ) = 4.00 g/mol
Substituting into the equation, you get \( n = \frac{0.486}{4.00} = 0.1215 \ \text{moles} \). Knowing the number of moles is critical for further calculations involving gas laws.
Pressure Conversion
Converting pressure units is crucial when working with the Ideal Gas Law, as different scenarios might require different units of measurement. In this exercise, we need to know the pressure in both pascals and atmospheres.

Once you calculate the pressure using the Ideal Gas Law (\( PV = nRT \)), you might obtain your results in pascals (Pa), the SI unit for pressure. However, often atmospheric pressure (atm) is used, especially in practical applications.
  • The conversion factor is: \( 1 \ \text{atm} = 101325 \ \text{Pa} \)
To convert from pascals into atmospheres, divide your pressure result in pascals by 101325. For instance, with a calculated pressure of 1477.3 Pa, simply compute \( \frac{1477.3}{101325} = 0.0146 \ \text{atm} \). This shows the pressure in a more commonly used unit.
Temperature Conversion
Working with gases often requires entering temperature in Kelvin into formulas, rather than Celsius, since Kelvin is the absolute temperature scale used in scientific calculations. To convert Celsius to Kelvin, use the simple formula: K = °C + 273.15.

In the given problem, the helium is at a temperature of 18.0°C. To find the corresponding temperature in Kelvin, simply add:
  • Temperature in Celsius = 18.0°C
  • Convert to Kelvin: \( T = 18 + 273.15 = 291.15 \ \text{K} \)
Kelvin temperatures ensure accuracy in gas calculations, as they start from absolute zero where theoretically all molecular motion stops. This conversion is pivotal when applying equations like the Ideal Gas Law.

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Most popular questions from this chapter

If deep-sea divers rise to the surface too quickly, nitrogen bubbles in their blood can expand and prove fatal. This phenomenon is known as the \(bends\). If a scuba diver rises quickly from a depth of 25 m in Lake Michigan (which is fresh water), what will be the volume at the surface of an N\(_2\) bubble that occupied 1.0 mm\(^3\) in his blood at the lower depth? Does it seem that this difference is large enough to be a problem? (Assume that the pressure difference is due to only the changing water pressure, not to any temperature difference. This assumption is reasonable, since we are warm-blooded creatures.)

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