Answer:
[tex]V_2= 736mL[/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 combined gas law:
[tex]\frac{P_2V_2}{T_2}= \frac{P_1V_1}{T_1}[/tex]
Thus, we solve for the final volume by solving for V2 as follows:
[tex]V_2= \frac{P_1V_1T_2}{T_1P_2}[/tex]
Now, we plug in the variables to obtain the result in milliliters and making sure we have both temperatures in Kelvins:
[tex]V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}\\\\V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}=736mL[/tex]
Regards!
