Question

A compressed cylinder of gas contains \(2.74 \times 10^{3} \mathrm{g}\) of \(\mathrm{N}_{2}\) gas at a pressure of \(3.75 \times 10^{7} \mathrm{Pa}\) and a temperature of \(18.7^{\circ} \mathrm{C} .\) What volume of gas has been released into the atmosphere if the final pressure in the cylinder is \(1.80 \times 10^{5}\) Pa? Assume ideal behavior and that the gas temperature is unchanged.

Short answer

The volume of gas released into the atmosphere is approximately \(16.15 \ \mathrm{m}^3\).

Step-by-step solution

Understand the Ideal Gas Law

The Ideal Gas Law is represented by the equation: \[PV = nRT\] Where: - P is the pressure of the gas (in Pascals) - V is the volume of the gas (in cubic meters) - n is the amount of substance (in moles) - R is the universal gas constant (\(8.314 \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)) - T is the temperature, in Kelvin (K)

Calculate the number of moles

First, we need to find the number of moles (n) of the N₂ gas. We can do this by using the molar mass of N₂ gas, which is \(28.02 \ \mathrm{g} \cdot \mathrm{mol}^{-1}\). The given mass of the gas is \(2.74 \times 10^3 \mathrm{g}\), so we can calculate the number of moles as follows: \[n = \frac{2.74 \times 10^3 \mathrm{g}}{28.02 \ \mathrm{g} \cdot \mathrm{mol}^{-1}}\] \(n \approx 97.79 \ \mathrm{mol}\)

Convert the temperature to Kelvin

The temperature is given in Celsius, but we need to convert it to Kelvin for the Ideal Gas Law. The conversion is as follows: \[T_\mathrm{K} = T_\mathrm{C} + 273.15\] So, \[T_\mathrm{K} = 18.7^{\circ} \mathrm{C} + 273.15\] \(T_\mathrm{K} \approx 291.85 \ \mathrm{K}\)

Calculate the initial volume

Use the Ideal Gas Law to find the initial volume (V₁) of the gas in the compressed cylinder: \[V_1 = \frac{nRT}{P_1}\] Where: - \(P_1 = 3.75 \times 10^7 \ \mathrm{Pa}\) (initial pressure) - R and n have been calculated in previous steps - T is in Kelvin So, \[V_1 = \frac{(97.79 \ \mathrm{mol})(8.314 \ \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1})(291.85 \ \mathrm{K})}{3.75 \times 10^7 \ \mathrm{Pa}}\] \(V_1 \approx 0.082 \ \mathrm{m}^3\)

Calculate the final volume

Use the Ideal Gas Law again to find the final volume (V₂) of the gas in the cylinder after the gas has been released: \[V_2 = \frac{nRT}{P_2}\] Where: - \(P_2 = 1.80 \times 10^5 \ \mathrm{Pa}\) (final pressure) So, \[V_2 = \frac{(97.79 \ \mathrm{mol})(8.314 \ \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1})(291.85 \ \mathrm{K})}{1.80 \times 10^5 \ \mathrm{Pa}}\] \(V_2 \approx 16.23 \ \mathrm{m}^3\)

Calculate the volume of gas released

Finally, subtract the initial volume (V₁) from the final volume (V₂) to find the volume of the gas (ΔV) that has been released into the atmosphere: \[\Delta V = V_2 - V_1\] So, \[\Delta V = 16.23 \ \mathrm{m}^3 - 0.082 \ \mathrm{m}^3\] \(\Delta V \approx 16.15 \ \mathrm{m}^3\) So, the volume of gas released into the atmosphere is approximately 16.15 cubic meters.

Key concepts

Moles Calculation

When dealing with gases in chemistry, calculating the number of moles is a fundamental step. Moles measure the amount of substance present, which is a crucial concept in understanding gas reactions and behaviors. In this exercise, we need to determine the moles of nitrogen gas (N₂) contained in a cylinder. The process is straightforward once we know the mass and molar mass of the gas.
The molar mass of nitrogen gas (N₂) is 28.02 g/mol. We're given a mass of 2.74 x 10³ g. To find the moles, use the formula:\[ n = \frac{\text{mass of gas}}{\text{molar mass}} \]Substituting the given values:\[ n = \frac{2.74 \times 10^3 \text{ g}}{28.02 \text{ g/mol}} \]This calculation provides approximately 97.79 moles of nitrogen gas.

• Moles help in understanding the quantity of gas present and are central to using the Ideal Gas Law.
• Remember to use molar mass appropriate for the gas in question.

Temperature Conversion to Kelvin

In the Ideal Gas Law, temperature must be in Kelvin. Kelvin is the SI unit of temperature and starts at absolute zero, where all molecular motion ceases. This conversion is a crucial detail when using the equation, \( PV = nRT \).
The given temperature in Celsius needs to be converted to Kelvin. The conversion formula is:\[ T_{\text{K}} = T_{\text{C}} + 273.15 \] For a temperature of 18.7°C:\[ T_{\text{K}} = 18.7 + 273.15 = 291.85 \text{ K} \]This ensures the Ideal Gas Law can be applied correctly.

• This conversion standardizes temperature measurements for gas calculations.
• Absolute zero in Kelvin is 0 K, equivalent to -273.15°C.

Volume Calculation

The calculation of volume changes in the gas within the cylinder is key in this exercise. Using the Ideal Gas Law, we can calculate the initial and final volumes of gas in the cylinder.

To find the initial volume \( V_1 \), we rearrange the Ideal Gas Law to solve for volume:\[ V_1 = \frac{nRT}{P_1} \]Given:

• Amount of substance \( n = 97.79 \text{ mol} \)
• Universal gas constant \( R = 8.314 \text{ J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)
• Temperature \( T = 291.85 \text{ K} \)
• Initial pressure \( P_1 = 3.75 \times 10^7 \text{ Pa} \)

This results in an initial volume:\[ V_1 \approx 0.082 \text{ m}^3 \]For the final volume \( V_2 \), the formula used is the same, but with the final pressure \( P_2 \):\[ V_2 = \frac{nRT}{P_2} \]With \( P_2 = 1.80 \times 10^5 \text{ Pa} \):\[ V_2 \approx 16.23 \text{ m}^3 \]Finally, to find the volume of gas released, subtract the initial volume from the final volume:\[ \Delta V = V_2 - V_1 \approx 16.15 \text{ m}^3 \]

• This calculation shows how volume changes under different pressure conditions.
• Gas volume is directly related to pressure, temperature, and moles present.