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Chapter 5: Problem 18

The volume occupied by \(9.0 \mathrm{~g}\) of nitrogen gas at \(300 \mathrm{~K}\) and \(750 \mathrm{mmHg}\) pressure is: (a) \(5.854\) (b) \(6.7432\) (c) \(8.8462\) (d) \(8.0225\)

Short Answer

Expert verified
The volume is \(8.0225\text{ L}\), option (d).

Step by step solution

01

Identify the Given Values

To solve the problem using the Ideal Gas Law, first identify the values provided: the mass of nitrogen gas is given as \(9.0 \text{ g}\), the temperature \(T\) is \(300 \text{ K}\), and the pressure \(P\) is \(750 \text{ mmHg}\).
02

Convert Pressure to Atmospheres

The Ideal Gas Constant \(R\) is commonly used with pressure in atmospheres, temperature in Kelvin, and volume in liters. First, convert the pressure from \(750 \text{ mmHg}\) to atmospheres using the conversion \(1 \text{ atm} = 760 \text{ mmHg}\).\[P = \frac{750}{760} = 0.9868 \text{ atm}\]
03

Calculate the Number of Moles

Use the molecular weight of nitrogen, which is \(28.02 \text{ g/mol}\). Calculate the number of moles \(n\) of nitrogen gas:\[n = \frac{9.0 \text{ g}}{28.02 \text{ g/mol}} = 0.3213 \text{ mol}\]
04

Use the Ideal Gas Law to Calculate Volume

The Ideal Gas Law is \(PV = nRT\). Solve for \(V\):\[V = \frac{nRT}{P}\]Substitute the known values into the equation, using \(R = 0.0821 \text{ L atm/mol K}\):\[V = \frac{0.3213 \times 0.0821 \times 300}{0.9868} \V \approx 8.0225 \text{ L}\]
05

Select Correct Answer from Options

Compare our calculated volume to the given options:- (a) \(5.854\)- (b) \(6.7432\)- (c) \(8.8462\)- (d) \(8.0225\)The correct answer is option (d) \(8.0225\).

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Key Concepts

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

Understanding the Molecular Weight of Nitrogen
The molecular weight, also known as molecular mass, of a substance is the sum of the atomic masses of all the atoms present in a single molecule of that substance. Understanding the molecular weight is crucial for many calculations in chemistry, especially when it involves converting between grams and moles.

For nitrogen gas, which exists as a diatomic molecule, we need to calculate the molecular weight of \(N_2\).

  • The atomic mass of a single nitrogen atom \(N\) is approximately \(14.01 \ ext{ amu}\).
  • As nitrogen gas is diatomic, we multiply this by 2 (since \(N_2\) consists of two nitrogen atoms).
  • This results in a molecular weight of approximately \(28.02 \ ext{ g/mol}\).
This information is used directly to convert the mass of nitrogen gas from grams to moles, which is a common step in gas-related calculations using the Ideal Gas Law.Determining moles is essential for applying chemical laws like the Ideal Gas Law, as it relates directly to the quantity of substance.
Conversion of Pressure Units
Pressure is a vital component in the Ideal Gas Law, and it's frequently measured in different units depending on the context.Common pressure units include atmospheres (atm), millimeters of mercury (mmHg), torr, and pascals (Pa).

In our specific problem, we start with a pressure in \(750 \text{ mmHg}\), but the Ideal Gas Law requires pressure to be in atmospheres when using the gas constant \(R = 0.0821 \text{ L atm/mol K}\).To make this conversion, we use the relationship between these two units:

  • 1 atmosphere ( ext{atm}) is equivalent to 760 millimeters of mercury ( ext{mmHg}).
  • To convert from ext{mmHg} to ext{atm}, divide the ext{mmHg} value by 760.
  • For example, \(P = \frac{750}{760} \approx 0.9868 \text{ atm}\).
This simple conversion ensures that the pressure is in the correct unit for further calculations with the Ideal Gas Law.Understanding unit conversions is a critical skill in chemistry that helps ensure the accuracy of computations in various chemical equations and relationships.
Calculation of Moles
Moles are units to measure the amount of substance in chemistry. This concept is fundamental in decoding the relationships between different substances in reactions and in finding out how much of something is involved in given conditions.

To calculate the number of moles (), we need two pieces of information:

  • The mass of the substance (in grams).
  • The molecular weight of the substance (in g/mol).
In our problem, we had \(9.0 \text{ g}\) of nitrogen gas, and as previously seen, nitrogen has a molecular weight of \(28.02 \text{ g/mol}\).

Calculating the moles involves dividing the mass by the molecular weight:\[ n = \frac{9.0 \text{ g}}{28.02 \text{ g/mol}} \approx 0.3213 \text{ mol} \]This value, known as the number of moles, is core to using the Ideal Gas Law equation \(PV = nRT\).It helps in establishing how the gas will behave under different conditions, linking directly to the volume when considering temperature and pressure.Knowing how to calculate moles efficiently will enhance your understanding of chemical properties and reactions.

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Most popular questions from this chapter

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