If 13.0 g of CO\(_2\) gas has a volume of 0.46 L at 310 K, what is its pressure in millimeters of mercury?

To find the pressure of CO2 gas in millimeters of mercury (mmHg), we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, convert the mass of CO2 to moles. The molar mass of CO2 is 44.01 g/mol. So, 13.0 g of CO2 is equal to (13.0 g)/(44.01 g/mol) = 0.295 mol.
Next, rearrange the ideal gas law equation to solve for pressure: P = (nRT)/V. Substitute the known values: P = (0.295 mol)(0.0821 L.mmHg/mol.K)(310 K)/(0.46 L) = 5.92 mmHg.

Answered by PhilDavis | 2024-06-18

Using the ideal gas law, the pressure of 13.0 g of CO₂ gas at 310 K and a volume of 0.46 L is approximately 5.92 mmHg. This was calculated by finding the number of moles of CO₂ and substituting all values into the rearranged ideal gas equation. The pressure was derived using the correct gas constant to yield the final units of mmHg.
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Answered by PhilDavis | 2024-12-26