p = 2.00 atm = 2026.50hPa T = 27°C = 300K m = 14.0g | Mr = 28g/mol | n = m/M = 14g/28g/mol = 0,5mol R = 83.1 [hPa L]/[mol K]
pV = nRT
nRT 0,5mol * 83.1[hPa L]/[mol K] * 300K V = ---------- = ----------------------------------------------------------- = 6.15L p 2026.5hPa
Answer B.
The** volume **of 14.0 g of nitrogen gas at a pressure of 2.00 atm and temperature of 27.0 degrees Celsius is 6.15 L. ;
The volume of 14.0 g of nitrogen gas at 2.00 atm and 27.0 °C is approximately 6.15 L, calculated using the ideal gas law. The steps included converting temperature to Kelvin, calculating the number of moles of nitrogen, and rearranging the ideal gas equation. The correct answer is B. 6.15 L.
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