Question

When 9.3 g of neon is combined in a 22 L container at 131^oC with 9.36 g of argon, what is the total pressure in mmHg?

Answer 1

Here, 9.3g of Ne is combined with 9.36g of Ar at 131°C temperature. So, number of moles of Ne present = (given mass/molar mass) = (9.3/20.18) = 0.46 moles. Similarly, number of moles of Ar present = (given mass/molar mass) = (9.36/39.948) = 0.23 moles. Now, we know the equation of state as, pV = nRT where, p is pressure, V is volume, n is number of moles, R is universal gas constant (0.082L-atm/mole K), T is the temperature (131°C = 273+131 = 404K). So, by using Dalton's law of partial pressure for a mixture of two gases, we can have p = (n₁RT/V) + (n₂RT/V) or, p_total = (n_totalRT/V) = {(0.46+0.23)×0.082×404}/22 = 1.039atm. Now, 1atm = 760mmHg, so 1.039atm = 789.65mmHg.

So, total pressure of the container would be 789.65mmHg.