Final answer:
Using the ideal gas law, the temperature of a 27.4 g sample of nitrogen gas (N₂) with a volume of 50.0 L and a pressure of 680 mmHg is calculated to be approximately 284 °C after converting the pressure to atm and the temperature to degrees Celsius from Kelvin.
Step-by-step explanation:
To find the temperature of a 27.4 g sample of nitrogen gas (N₂) with a volume of 50.0 L and a pressure of 680 mmHg, we must use the ideal gas law, which is PV = nRT. Here, P is the pressure of the gas in atmospheres (atm), V is the volume in liters (L), n is the number of moles of the gas, R is the ideal gas constant (0.0821 L.atm.mol⁻¹.K⁻¹), and T is the temperature in Kelvin (K).
First, we convert the pressure from mmHg to atm by using the conversion factor 760 mmHg = 1 atm, which gives us a pressure of approximately 0.895 atm (680 mmHg / 760 mmHg = 0.895 atm). The molar mass of molecular nitrogen, N₂, is 28.01 g/mol. We then calculate the number of moles of N₂ which is 27.4 g / 28.01 g/mol ≈ 0.978 moles.
Once we have all the values, we can rearrange the ideal gas law to solve for T: T = (PV)/(nR). Substituting the values, we find T in Kelvin:
T = (0.895 atm * 50.0 L) / (0.978 moles * 0.0821 L.atm.mol⁻¹.K⁻¹) ≈ 557 K.
To convert the temperature to degrees Celsius, we use the formula °C = K - 273.15. Therefore, the temperature of the gas is 557 K - 273.15 K ≈ 284 °C.