What is the volume, in liters, occupied by a mixture of 15.2 \(\mathrm{g} \mathrm{Ne}(\mathrm{g})\) and \(34.8 \mathrm{g} \mathrm{Ar}(\mathrm{g})\) at 7.15 atm pressure and \(26.7^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The volume of the gas mixture is approximately 5.45 L

Step by step solution

01

Convert the masses of the gases to moles

The molar mass of Neon is 20.2 g/mol and that of Argon is 39.95 g/mol. So, the number of moles of Neon can be calculated as \(15.2 \, g \times \frac{1 mol}{20.2 \, g} = 0.75 \, mol\), and the number of moles of Argon can be calculated as \(34.8 \, g \times \frac{1 mol}{39.95 \, g} = 0.87 \, mol\).
02

Sum the total moles and convert temperature

Add the moles of Neon and Argon to get the total moles of gases. So \(n = 0.75 \, mol + 0.87 \, mol = 1.62 \, mol\). Convert the temperature from Celsius to Kelvin by adding 273.15 (since \(K = ^\circ C + 273.15\)). So \(T = 26.7 + 273.15 = 299.85 \, K\).
03

Calculate the volume using the Ideal Gas Law

Substitute the values you know into the equation PV = nRT and solve for V. The gas constant, R, when pressure is given in atm and volume in litres is \(0.0821 \, L.atm/K.mol\). So \(V = \frac{nRT}{P} = \frac{1.62 \, mol \times 0.0821 \, L.atm/K.mol \times 299.85 \, K}{7.15 \, atm}\)

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