Answer:
620.71 L the final volume of the balloon.
Step-by-step explanation:
Initial volume of the gas in the balloon=
![V_1=3.50 L](https://img.qammunity.org/2020/formulas/chemistry/high-school/tjimvdt5t4yfq8ojex9na1zx5e4ivkq32y.png)
Initial pressure of the gas in the balloon=
![P_1=1.20 atm](https://img.qammunity.org/2020/formulas/chemistry/high-school/arplpnt02cdmtpztebgxw7o8egroxiix4a.png)
Initial temperature of the gas in the balloon=
![T_1=18^oC =291.15 K](https://img.qammunity.org/2020/formulas/chemistry/high-school/gs30xn0x10svo023x8hftae78d2hvx9fx4.png)
Moles of gases = n
...[1]
Final volume of the gas in the balloon =
![V_2=3.50 L](https://img.qammunity.org/2020/formulas/chemistry/high-school/f5og95eonipcl4j7x24u9kiomidhijnag8.png)
Final pressure of the gas in the balloon =
![P_2=5.70* 10^(-3) atm](https://img.qammunity.org/2020/formulas/chemistry/high-school/kkbn0f1cps5x2e6kw9o2cz6jb4q3h1xgyl.png)
Final temperature of the gas in the balloon =
![T_2=-45^oC =228.15 K](https://img.qammunity.org/2020/formulas/chemistry/high-school/g1j0lua6il6ze4p9cx2zuilmxhjccwklgp.png)
Moles of gases = n
...[2]
[1] = [2]
![(P_2V_2)/(RT_2)=(P_1V_1)/(RT_1)](https://img.qammunity.org/2020/formulas/chemistry/high-school/1t9c41guc0196irxcy6tzqy0f636gt59jf.png)
![V_2=(P_1V_1* T_2)/(T_1* P_2)](https://img.qammunity.org/2020/formulas/chemistry/high-school/8895aywwasq0u3xz20l3n8jt8i0yvvn353.png)
![V_2=(1.29 atm * 3.50 L* 228.15 K)/(5.70* 10^(-3) atm* 291.15 K)=620.71 L](https://img.qammunity.org/2020/formulas/chemistry/high-school/3tz47j32gk1b85g9nb7eggzfju1d74dlxv.png)
620.71 L the final volume of the balloon.