A sample of He gas (V = 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 in a 9.0 L flask (which was initially evacuated). What is the total pressure in the 9.0 L flask after mixing? Assume the temperature remains at 25 °C.

Question

Chemistry

Alexzander Miller

14 March, 04:49

A sample of He gas (V = 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 in a 9.0 L flask (which was initially evacuated). What is the total pressure in the 9.0 L flask after mixing? Assume the temperature remains at 25 °C.

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Answers (1)

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Darien Richard · Answer 1

3.67 atm

Explanation:

Because the gases are not reacting and the temperature remaining the same, we can apply Boyle's Law:

P1V1 = P2V2

Where P is the pressure, V is the volume, 1 is the initial state, and 2 the final state.

For a mixture of gases, from Dalton's law, the total pressure is the sum of the partial pressures of the components, so:

P1V1 = PHe*Ve + PNe*VNe

PHe*Ve + PNe*VNe = P2*V2

5.6*3.0 + 3.6*4.5 = P2*9

9P2 = 33

P2 = 3.67 atm