Question

Liquid \(\mathrm{N}_{2}\) has a density of \(875.4 \mathrm{kg} \mathrm{m}^{-3}\) at its normal boiling point. What volume does a balloon occupy at \(298 \mathrm{K}\) and a pressure of 1.00 atm if \(3.10 \times 10^{-3} \mathrm{L}\) of liquid \(\mathrm{N}_{2}\) is injected into it? Assume that there is no pressure difference between the inside and outside of the balloon.

Short answer

First, find the mass of injected liquid nitrogen using the volume and density: \(m = (875.4 \mathrm{kg}\mathrm{m}^{-3}) \times (3.10 \times 10^{-3} \mathrm{L} \times 10^{-3} \mathrm{m}^3\mathrm{L}^{-1})\). Next, calculate the number of moles (n) using the mass and molecular weight of nitrogen: \(n = \frac{m \times 10^3 \mathrm{g}\mathrm{kg}^{-1}}{28.0 \mathrm{g}\mathrm{mol}^{-1}}\). Finally, use the ideal gas law \(PV = nRT\) to find the volume of the balloon at the given conditions, with R as the ideal gas constant, T as 298 K, and P as 1.00 atm converted to Pascal (101325 Pa): \(V = \frac{nRT}{P}\). Convert the resulting volume from m³ to liters by multiplying by 1000.

Step-by-step solution

Determine the mass of injected liquid nitrogen

Given that the volume of liquid nitrogen injected is \(3.10 \times 10^{-3} \mathrm{L}\), we can calculate the mass of the injected liquid nitrogen: Density \(= \frac{mass}{volume}\) Mass \(= \text{density} \times \text{volume}\) We are given the density of liquid nitrogen to be \(875.4 \mathrm{kg}\mathrm{m}^{-3}\). To convert the volume to m³, multiply by \(10^{-3}\). Thus, the mass (m) can be calculated as: \(m = (875.4 \mathrm{kg}\mathrm{m}^{-3}) \times (3.10 \times 10^{-3} \mathrm{L} \times 10^{-3} \mathrm{m}^3\mathrm{L}^{-1})\)

Calculate the number of moles of injected liquid nitrogen

The next step is to calculate the number of moles (n) of the injected liquid nitrogen using the molecular weight (M) of nitrogen, \(M(\mathrm{N}_{2}) = 28.0 \mathrm{g}\mathrm{mol}^{-1}\): Number of moles, \(n = \frac{mass}{molecular\ weight}\) First, we must convert the mass into grams before calculating the number of moles. Thus: \(n = \frac{m \times 10^3 \mathrm{g}\mathrm{kg}^{-1}}{M(\mathrm{N}_{2})}\)

Use the ideal gas law to find the volume of the balloon

Now that we have the number of moles of nitrogen, we can use the ideal gas law equation to find the volume (V) of the balloon at the given conditions: Ideal gas law, \(PV = nRT\) Here, R is the ideal gas constant, \(R = 8.314 \mathrm{J\ K}^{-1}\mathrm{mol}^{-1}\), T is the temperature in Kelvin, and P is the pressure in Pascal (Pa). We are given T as 298 K and P as 1.00 atm. However, the pressure must be converted to Pascal: 1 atm = 101325 Pa We can now calculate the volume of the balloon by plugging in the values: \(V = \frac{nRT}{P}\) After solving the equation, we will have the volume of the balloon in m³. This value should be converted back to liters by multiplying by 1000.

Key concepts

Density Calculation

Density is a key concept that helps us understand how much mass is present in a given volume of a substance. In simpler terms, it tells us how "packed" or "concentrated" the matter is within a certain space. The formula for density is:

• Density = \( \frac{\text{mass}}{\text{volume}} \)

By rearranging this formula, we can also calculate mass if the density and volume are known. This was used in the exercise where the mass of liquid nitrogen was needed. The known density of liquid nitrogen, \(875.4 \, \text{kg m}^{-3}\), allowed for the mass calculation by multiplying it with the volume of liquid nitrogen injected.
For volume, it's crucial to ensure the units match. The volume of liquid nitrogen was provided in liters, so it required conversion to cubic meters (m³) to match the density units. Thus, you multiply the volume in liters by \(10^{-3}\) to convert to m³, making the calculations seamless.

Molar Mass

Molar mass, often seen as molecular weight, is a measurement used to quantify how many grams one mole of a substance weighs. It connects the tiny atomic scale of chemical substances to the macroscopic scale we can measure in the laboratory.
For nitrogen gas (\(\mathrm{N}_{2}\)), the molar mass is \(28.0 \, \text{g mol}^{-1}\). Understanding molar mass allows us to convert between mass and moles, which is critical for many calculations in chemistry, including those involving the Ideal Gas Law.

• To find the number of moles, use: \( n = \frac{\text{mass}}{\text{molar mass}} \)

In the original exercise, after calculating the mass of nitrogen, this step enabled converting the mass, initially found in kilograms, into moles. Remember, converting the mass into grams (from kilograms) is essential, as molar mass is typically in grams per mole. This aligns the units for correct computation of moles.

Unit Conversion

Unit conversion is a fundamental aspect of working with scientific data. It ensures all measurements are consistent for accurate calculations. In the context of the Ideal Gas Law and other physics applications, it is crucial that units align.
In this exercise, several conversions were necessary:

• The volume of liquid nitrogen was converted from liters to cubic meters (\( \text{L} \to \text{m}^3 \) by multiplying by \(10^{-3}\)).
• The pressure was converted from atmospheres (atm) to Pascals (Pa). The conversion factor is: \(1\, \text{atm} = 101,325\, \text{Pa}\).
• The mass of the nitrogen was also converted from kilograms to grams to compute moles using molar mass.

By maintaining consistent and proper units, calculations become straightforward, allowing the accurate application of equations like the Ideal Gas Law: \( PV = nRT \). It and similar equations require particular units, making conversions a standard step in problem-solving.