Question

1. consider the three flasks in the diagram below. assuming the connecting tubes have negligible volume, what is the partial pressure of each gas and the total pressure after all the stop - cocks are opened? 1.00 l 200 torr he 1.00 l 0.400 atm ne 2.00 l 24.0 kpa ar

Explanation:

Step1: Convert all pressures to the same unit (kPa)

1 torr = 0.133322 kPa, 1 atm = 101.325 kPa.
For He: $P_{He1}=200\ torr\times0.133322\ kPa/torr = 26.6644\ kPa$.
For Ne: $P_{Ne1}=0.400\ atm\times101.325\ kPa/atm = 40.53\ kPa$.
For Ar: $P_{Ar1}=24.0\ kPa$.

Step2: Calculate the total volume

$V_{total}=1.00\ L + 1.00\ L+2.00\ L = 4.00\ L$.

Step3: Use Boyle's law ($P_1V_1 = P_2V_2$) to find the partial - pressures after opening the stop - cocks

For He: $P_{He2}=\frac{P_{He1}V_{He1}}{V_{total}}=\frac{26.6644\ kPa\times1.00\ L}{4.00\ L}=6.67\ kPa$.
For Ne: $P_{Ne2}=\frac{P_{Ne1}V_{Ne1}}{V_{total}}=\frac{40.53\ kPa\times1.00\ L}{4.00\ L}=10.13\ kPa$.
For Ar: $P_{Ar2}=\frac{P_{Ar1}V_{Ar1}}{V_{total}}=\frac{24.0\ kPa\times2.00\ L}{4.00\ L}=12.0\ kPa$.

Step4: Calculate the total pressure

$P_{total}=P_{He2}+P_{Ne2}+P_{Ar2}=6.67\ kPa + 10.13\ kPa+12.0\ kPa = 28.8\ kPa$.

Answer:

The partial pressure of He is $6.67\ kPa$, the partial pressure of Ne is $10.13\ kPa$, the partial pressure of Ar is $12.0\ kPa$, and the total pressure is $28.8\ kPa$.