Question

A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in a 9.0 L flask. The total pressure in the flask was __________ atm. Assume the initial pressure in the flask was 0.00 atm and the temperature upon mixing was 25°C. Select one:

Answer 1

Answer:

$P=3.7atm$

Explanation:

Hello,

In this case, it is possible to determine the pressures of both helium and neon as shown below:

$n_{He}=\frac{P_{He}V_{He}}{RT}=\frac{5.6atm*3.0L}{0.082\frac{atm*L}{mol*K}*298.15K} =0.688molHe\\\\n_{Ne}=\frac{P_{Ne}V_{Ne}}{RT}=\frac{3.6atm*4.5L}{0.082\frac{atm*L}{mol*K}*298.15K}=0.663molNe$

Now, one considers the total moles (addition between both neon's and helium's moles) and the total volume to compute the final pressure as shown below:

$P=\frac{n_TRT}{V_T} =\frac{(0.688+0.663)mol*0.082\frac{atm*L}{mol*K}*298.15K}{9.0L}=3.7atm$

Best regards.