What is the volume,in L,occupied by a mixture of 16.0 g Ne(g)and 42.0 g Ar(g)at 15.0 atm pressure and 25 °C?

Question

Quizplus · Accepted Answer

The volume can be calculated using the ideal gas law, $PV = nRT$, 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 temperature to Kelvin: $25°C + 273.15 = 298.15 K$. Next, calculate the moles of Ne and Ar: Ne (20.18 g/mol) has $16.0 g / 20.18 g/mol = 0.793$ moles, and Ar (39.95 g/mol) has $42.0 g / 39.95 g/mol = 1.051$ moles. The total moles are $0.793 + 1.051 = 1.844$ moles. Using the ideal gas law, rearrange to solve for $V$: $V = \frac{nRT}{P}$. Substituting the values: $V = \frac{1.844 moles \times 0.0821 \frac{L \cdot atm}{mol \cdot K} \times 298.15 K}{15.0 atm}$, which calculates to approximately $3.01 L$.

What Is the Volume,in L,occupied by a Mixture of 16