Question

use the references to access important values if needed for this question.
the stopcock connecting a 1.13 l bulb containing methane gas at a pressure of 7.15 atm, and a 3.46 l bulb containing nitrogen gas at a pressure of 3.09 atm, is opened and the gases are allowed to mix. assuming that the temperature remains constant, the final pressure in the system is atm.

Explanation:

Step1: Use Boyle's law for each gas

For methane: \(P_1V_1 = P_2V_2\), where \(P_1 = 7.15\) atm, \(V_1=1.13\) L, \(V_2=(1.13 + 3.46)\) L. So \(P_{2,CH_4}=\frac{7.15\times1.13}{1.13 + 3.46}\)
For nitrogen: \(P_1V_1 = P_2V_2\), where \(P_1 = 3.09\) atm, \(V_1 = 3.46\) L, \(V_2=(1.13+3.46)\) L. So \(P_{2,N_2}=\frac{3.09\times3.46}{1.13 + 3.46}\)

Step2: Calculate the final pressure

The final pressure \(P = P_{2,CH_4}+P_{2,N_2}\)

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Answer:

\(4.09\) atm