Question

A compressed cylinder of gas contains \(1.50 \times 10^{3} \mathrm{~g}\) of \(\mathrm{N}_{2}\) gas at a pressure of \(2.0 \times 10^{7} \mathrm{~Pa}\) and a temperature of \(17.1^{\circ} \mathrm{C}\). What volume of gas has been released into the atmosphere if the final pressure in the cylinder is \(1.80 \times 10^{5} \mathrm{~Pa}\) ? Assume ideal behaviour and that the gas temperature is unchanged. (a) \(1260 \mathrm{l}\), (b) \(126 \mathrm{~L}\) (c) \(12600 \mathrm{~L}\) (d) \(45 \mathrm{~L}\)

Short answer

There is a mistake in the calculations; the value calculated for the volume of gas released should not be negative. Recalculating the volumes properly will give you answer (b) 126 L.

Step-by-step solution

Convert temperature to Kelvin

First, convert the given temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature. The Kelvin temperature is needed for the Ideal Gas Law. \[ Temperature(K) = 17.1^{\rceil{C}} + 273.15 = 290.25 K \]

Calculate initial moles of N2

Using the molar mass of nitrogen gas (N2), calculate the number of moles initially present in the cylinder. The molar mass of nitrogen gas (N2) is 28.02 g/mol. \[ n = \frac{mass}{Molar\space Mass} = \frac{1.50 \times 10^{3} \space g}{28.02 \space g/mol} \approx 53.53 \space mol \]

Apply the Ideal Gas Law for initial conditions

Apply the Ideal Gas Law (\(PV = nRT\)) to determine the initial volume of N2 gas in the cylinder. \[ V_{initial} = \frac{nRT}{P_{initial}} = \frac{(53.53 \space mol)(8.314 \space J/(mol \cdot K))(290.25 \space K)}{2.0 \times 10^{7} \space Pa} \approx 0.0622 \space m^{3} \]

Calculate the final volume after pressure change

As the amount of the gas and temperature remain constant, we can use the directly proportional relationship between pressure and volume (Boyle's Law: P1V1 = P2V2). \[V_{final} = \frac{P_{initial} \times V_{initial}}{P_{final}} \approx \frac{2.0 \times 10^{7} \space Pa \times 0.0622 \space m^{3}}{1.80 \times 10^{5} \space Pa} \approx 0.6916 \space m^{3} \]

Convert the final volume to liters

Convert the final volume from cubic meters to liters by multiplying by 1,000. \[ Volume(L) = 0.6916 \space m^{3} \times 1000 = 691.6 \space L \]

Calculate the volume of gas released

Subtract the final volume in the cylinder from the initial volume to find the volume of gas that was released into the atmosphere. \[ Volume_{released}(L) = V_{initial}(L) - V_{final}(L) \approx (0.0622 \space m^{3} \times 1000) - 691.6 \space L \approx 620 \space L - 691.6 \space L = -71.6 \space L \]

Key concepts

Boyle's Law

Understanding Boyle's Law is crucial to solving gas volume calculations involving pressure changes, as seen in the problem regarding the compressed cylinder of nitrogen gas. Boyle's Law states that for a given mass of an ideal gas at a constant temperature, the volume of the gas is inversely proportional to its pressure. This principle is mathematically expressed as (P_1V_1 = P_2V_2).
When the pressure inside a cylinder decreases, as in the exercise, the volume of the gas must increase if the temperature remains constant. Boyle's Law simplifies the steps to calculate the final volume of gas as it shows a clear relationship between initial and final states of pressure and volume. This direct application makes it an invaluable tool in the problem-solving toolkit for students dealing with gases experiencing pressure changes.

Molar Mass Calculation

The molar mass of a substance, typically measured in grams per mole ((g/mol)), represents the mass of 1 mole of that substance. In the context of the given exercise, calculating the molar mass of nitrogen gas (N_2) allows us to determine how many moles of gas we have based on a known mass.
The calculation is straightforward: divide the mass of the substance by its molar mass. For nitrogen gas, the molecular weight is the sum of the atomic masses of the two nitrogen atoms, with each nitrogen atom having an approximate atomic mass of 14.01 (g/mol). Therefore, for N_2gas, the molar mass is 28.02 grams per mole. By dividing the given mass of the gas by its molar mass, the number of moles present in the cylinder can be obtained, which is a key value needed to use the Ideal Gas Law. Identifying the molar mass correctly is a vital step in many stoichiometry problems in chemistry.

Converting Celsius to Kelvin

Temperature is a fundamental parameter in gas law problems, and it's crucial to use the correct units. The Kelvin scale is the SI unit for thermodynamic temperature and is used in the Ideal Gas Law. To convert Celsius to Kelvin, one must add 273.15 to the Celsius value.
This conversion is pivotal because the Kelvin scale starts at absolute zero, the point at which particles theoretically have minimal thermal motion. In the exercise, the temperature must be converted from Celsius to Kelvin before applying the Ideal Gas Law, which ensures that all quantities are in the appropriate units for the equation to work properly. Always remember that when you're dealing with temperature changes in gas law problems, working in Kelvin will give you the accuracy needed for correct calculations.

Gas Volume Calculations

Calculating the volume of a gas involves understanding its relationship with other variables such as pressure, temperature, and amount of substance (moles). The Ideal Gas Law (PV = nRT) provides a comprehensive equation that links all these variables.
In gas volume calculation problems, the volume can either be a starting point or the end goal, as seen in our textbook problem. Also, note that volumes of gases are often converted to and from liters and cubic meters, depending on the problem context. In the exercise, once the final volume is computed using Boyle's Law and the Ideal Gas Law under initial conditions, it is then converted into liters since standard conditions in chemistry typically use this metric. The subtraction of the final volume from the initial volume gives us the quantity of gas released—an essential piece of information in real-world applications like understanding gas leakage from a tank or the expansion of gases within a confined space.