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A 14.0 tank contains 250 g of methane (CH4) gas at 27 atm at 298 K. Accidentally, 190 g of CO2 was added to the tank. What will be the resulting pressure of the mixture in the tank? Assume that no CH4 leaked out as the CO2 gas waa being added.​

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Answer:

[tex]P=34.8atm[/tex]

Explanation:

Hello!

In this case, since the tank has a constant volume as well as temperature, we would be able to compute the resulting pressure via:

[tex]P=\frac{n_TRT}{V}[/tex]

Whereas the total moles are computed by adding the moles of each gas as shown below:

[tex]n_{CH_4}=250gCH_4*\frac{1molCH_4}{16.05gCH_4}=15.6molCH_4\\\\n_{CO_2}=190g CO_2*\frac{1molCO_2}{44.01gCO_2}=4.32molCO_2\\\\n_T= 15.6molCH_4+4.32molCO_2\\\\n_T=19.9mol[/tex]

Thus, we plug in the total pressure, temperature and volume to get the resulting pressure:

[tex]P=\frac{19.9mol*0.08206\frac{atm*L}{mol*K}*298K}{14.0L}\\\\P=34.8atm[/tex]

Best regards!

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