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An ideal gas at 2.0 atm pressure and 298 K temperature has a volume of 12.0 L. If the volume is decreased to 4.0 L and the temperature is held constant, what is the new pressure of the the gas in atm

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Sebassandinidn

Answer:

[tex]P_2=6.0atm[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the Boyle's law, since the temperature is constant and both volume and pressure change:

[tex]P_2V_2=P_1V_1[/tex]

Thus, we solve for the final pressure, P2, to obtain:

[tex]P_2=\frac{P_1V_1}{V_2}[/tex]

And plug in the given data to get:

[tex]P_2=\frac{2.0atm*12.0L}{4.0L}\\\\P_2=6.0atm[/tex]

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