Answer:
[tex]V_2=1228.9L[/tex]
Explanation:
Hello there!
In this case, given the pressure, temperature and volume of the gas, we notice that we need the combined ideal gas as shown below:
[tex]\frac{P_2V_2}{T_2} =\frac{P_1V_1}{T_1}[/tex]
Thus, solving for the final volume, V2, we would obtain:
[tex]V_2=\frac{P_1V_1T_2}{T_1P_2}[/tex]
Now, we plug in the data and make sure the temperature must be in Kelvins to obtain:
[tex]V_2=\frac{200atm*6.0L*(25+273)K}{(18+273.15)K(1.0atm)}\\\\V_2=1228.9L[/tex]
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