Question

What is the density of $N_2$ gas at $227°C$ and 5.00 atm. pressure ? (R = 0.082 L Atm $K^{-1} mol^{-1}$)

• A. 1.40 g/mL
• B. 2.81 g/mL
• C. 3.41 g/mL
• D. 0.29 g/mL

Answer 1

The Correct Option is C

To find the density of $N_2$ gas at $227^\circ C$ and 5.00 atm pressure, we use the ideal gas law formula and manipulate it to solve for density.

The ideal gas law is given by:

PV = nRT

Where:

• P is the pressure of the gas.
• V is the volume of the gas.
• n is the number of moles.
• R is the universal gas constant, 0.082 \, \text{L} \, \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1}.
• T is the temperature in Kelvin.

First, convert the temperature from Celsius to Kelvin:

T(K) = 227^\circ C + 273 = 500 \, K

We need to find density (\rho), which is mass per unit volume. We know:

\rho = \frac{m}{V}

From the ideal gas equation PV = nRT, and knowing that n = \frac{m}{M} (where M is the molar mass), we get:

PV = \frac{m}{M}RT

Rearranging for \frac{m}{V}, we obtain:

\frac{m}{V} = \frac{PM}{RT}

The molar mass of $N_2$ is 28 g/mol. Now, substitute the values:

\rho = \frac{5 \, \text{atm} \times 28 \, \text{g/mol}}{0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 500 \, K}

Calculating this gives:

\rho = \frac{140}{41} \approx 3.41 \, \text{g/L}

Therefore, the density of $N_2$ gas under the given conditions is approximately 3.41 \, \text{g/L} or 3.41 \, \text{g/mL}.

Thus, the correct answer is 3.41 \, \text{g/mL}.