Answer:
[tex]V_2=11.6L[/tex]
Explanation:
Hello!
In this case, since we volume, pressure and temperature which are all changing, we can use the combined ideal gas law to write:
[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
Thus, since the final volume V2 is required, by solving for it, we write:
[tex]V_2=\frac{P_1V_1T_2}{T_1P_2}[/tex]
In such a way, we plug in the given data to obtain:
[tex]V_2=\frac{3.36atm*15.0L*383K}{298K*5.60atm}\\\\V_2=11.6L[/tex]
Which means that the process compressed the gas.
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