Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations. Consider a 1000-megawatt coal-fired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, $101.3 \mathrm{kPa}\(, and \)27^{\circ} \mathrm{C}$, calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. (b) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and $12.16 \mathrm{MPa}\( and a density of \)1.2 \mathrm{~g} / \mathrm{cm}^{3},$ what volume does it possess? (c) If it is stored underground as a gas at \(30^{\circ} \mathrm{C}\) and \(7.09 \mathrm{MPa}\), what volume does it occupy?

Step by step solution

01

Convert the mass of CO2 from tons to grams

: Given, the power plant produces \(6 \times 10^6\) tons of CO2 per year. We need to convert it into grams: \(1 \: \text{ton} = 1000 \: \text{kg}\) and \(1 \: \text{kg} = 1000 \: \text{g}\). So, the mass of CO2 in grams is: \(6 \times 10^6 \times 1000 \times 1000 = 6 \times 10^{12} \: \text{g}\).

02

Calculate the number of moles of CO2

: The molecular weight of CO2 is approximately \(44 \: \text{g/mol}\). We will use this to calculate the number of moles of CO2 produced: Number of moles = \( \frac{6 \times 10^{12}}{44} \: \text{moles}\).

03

Calculate the volume of CO2 at given conditions for part (a)

: For part (a), we are given the conditions as \(101.3 \: \text{kPa}\) pressure and \(27^{\circ} \mathrm{C}\). We will first convert the temperature to Kelvin: \(T = 27 + 273.15 = 300.15 \: \mathrm{K}\). Now, we will use the ideal gas law formula: \(PV = nRT\), where P is the pressure, V is the volume of the gas, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. We can rearrange the formula to solve for V: \(V = \frac{nRT}{P}\). Substituting the values for temperature, pressure, and the universal gas constant (R = 8.314 J/mol·K) into the equation, we can calculate the volume of CO2 produced under part (a) conditions: \(V = \frac{(6 \times 10^{12}/44) \times 8.314 \times 300.15}{101.3 \times 10^3} \: \mathrm{m}^{3}\).

04

Calculate the volume for parts (b) and (c) using given densities

: In parts (b) and (c), the density of CO2 is given, and we are asked to find the volume that it occupies. To find the volume, we can use the formula: Volume = \(\frac{\text{mass}}{\text{density}}\). For part (b), the mass of CO2 is \(6 \times 10^{12}\) g, and its density is \(1.2 \: \mathrm{g}/ \mathrm{cm}^{3}\). Volume = \(\frac{6 \times 10^{12}}{1.2 \: \mathrm{g}/\mathrm{cm}^{3}} \: \mathrm{cm}^{3}\). For part (c), we are given a density of \(2.45 \: \text{g/cm}^{3}\). Volume = \(\frac{6 \times 10^{12}}{2.45 \: \mathrm{g}/\mathrm{cm}^{3}} \: \mathrm{cm}^{3}\). Note that the final volumes should be expressed in appropriate units (e.g., \(\mathrm{m}^{3}\) or \(\mathrm{cm}^{3}\)) and converted if necessary.

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