Answer:
[tex]V_2=2487.9L[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the final volume by using the combined ideal gas:
[tex]\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}[/tex]
In such a way, by solving for the final volume, V2, we obtain:
[tex]V_2=\frac{P_1V_1T_2}{P_2T_1} \\\\[/tex]
Now, by plugging in the pressures, temperatures in Kelvins and initial volume, we will obtain:
[tex]V_2=\frac{(400.0mmHg)(120.0L)(283.15K)}{(20.0mmHg)(273.15K)}\\\\V_2=2487.9L[/tex]
Regards!