Question

What will be the density of \( N_2 \) gas at 230°C and 3 atm pressure? (Given \( R = 0.082 \, \text{L atm} \, \text{mol}^{-1} \, \text{K}^{-1} \))

• A. 3.41 g/mL
• B. 2.03 g/mL
• C. 4.30 g/mL
• D. 0.27 g/mL

Hint:

When calculating density from the ideal gas law, ensure to convert all units to consistent SI units and use the molar mass of the gas.

Answer 1

The Correct Option is B

The ideal gas equation is given by: \[ PV = nRT \] Where: - \( P \) = pressure = 3 atm - \( V \) = volume of 1 mole of gas (at conditions of 1 mol) - \( n \) = number of moles = 1 mol - \( R \) = ideal gas constant = 0.082 \, L atm mol\(^{-1}\) K\(^{-1}\) - \( T \) = temperature in Kelvin = 230°C + 273 = 503 K Now, rearrange the equation to solve for volume of 1 mol gas: \[ V = \frac{nRT}{P} = \frac{(1 \, \text{mol})(0.082 \, \text{L atm} \, \text{mol}^{-1} \, \text{K}^{-1})(503 \, \text{K})}{3 \, \text{atm}} = \frac{41.216}{3} = 13.74 \, \text{L} \] The density \( \rho \) is given by: \[ \rho = \frac{\text{Mass}}{\text{Volume}} \] The molar mass of \( N_2 \) is 28 g/mol. So, the density is: \[ \rho = \frac{28 \, \text{g/mol}}{13.74 \, \text{L}} = 2.03 \, \text{g/L} = 2.03 \, \text{g/mL} \] Thus, the correct answer is 2.03 g/mL.