Question

three flasks, each containing a gas sample, are set up as shown below. assuming no temperature change, determine the final pressure inside the system after all stopcocks are opened. assume that the connecting tube has negligible volume.
a = he(g) v = 4.21 l p = 0.890 atm
b = kr(g) v = 2.47 l p = 1.14 atm
c = ch₄(g) v = 5.73 l p = 2.79 atm
final pressure = atm

Explanation:

Step1: Calculate the partial pressure of each gas using \(P_1V_1 = P_2V_2\)

The total volume \(V_{total}=4.21 + 2.47+5.73=12.41\space L\)
For gas \(A\) (\(He\)):
\(P_{A1}V_{A1}=P_{A2}V_{total}\)
\(P_{A2}=\frac{P_{A1}V_{A1}}{V_{total}}\)
Substitute \(P_{A1} = 0.890\space atm\), \(V_{A1}=4.21\space L\), \(V_{total}=12.41\space L\)
\(P_{A2}=\frac{0.890\times4.21}{12.41}\)
\(P_{A2}=\frac{3.7469}{12.41}\approx0.302\space atm\)

Step2: Calculate the partial pressure of gas \(B\) (\(Kr\))

Using \(P_{B1}V_{B1}=P_{B2}V_{total}\)
\(P_{B2}=\frac{P_{B1}V_{B1}}{V_{total}}\)
Substitute \(P_{B1} = 1.14\space atm\), \(V_{B1}=2.47\space L\), \(V_{total}=12.41\space L\)
\(P_{B2}=\frac{1.14\times2.47}{12.41}\)
\(P_{B2}=\frac{2.8158}{12.41}\approx0.227\space atm\)

Step3: Calculate the partial pressure of gas \(C\) (\(CH_4\))

Using \(P_{C1}V_{C1}=P_{C2}V_{total}\)
\(P_{C2}=\frac{P_{C1}V_{C1}}{V_{total}}\)
Substitute \(P_{C1} = 2.79\space atm\), \(V_{C1}=5.73\space L\), \(V_{total}=12.41\space L\)
\(P_{C2}=\frac{2.79\times5.73}{12.41}\)
\(P_{C2}=\frac{15.9867}{12.41}\approx1.29\space atm\)

Step4: Calculate the final pressure using Dalton's law of partial pressures \(P_{total}=P_{A2}+P_{B2}+P_{C2}\)

\(P_{total}=0.302 + 0.227+1.29\)
\(P_{total}=1.819\approx1.82\space atm\)

Answer:

\(1.82\)