Answer:
![\boxed {\boxed {\sf 4.6 \ Liters}}](https://img.qammunity.org/2022/formulas/chemistry/high-school/yhjwxrtlfsuuq6sayvrh2x8uopci3jtz41.png)
Step-by-step explanation:
The pressure stays constant, so we are dealing with volume and temperature, so we use Charles's Law. This states the temperature and volume of a gas are directly proportional. The formula is:
![(V_1)/(T_1)=(V_2)/(T_2)](https://img.qammunity.org/2022/formulas/chemistry/college/p0u7agjf5ksvdaya542rtohs6breeoi1pg.png)
We know the original balloon has a volume of 1.25 liters at a temperature of 23 degrees celsius. These values can be substituted in.
![(1.25 \ L)/(23 \textdegree C)=(V_2)/(T_2)](https://img.qammunity.org/2022/formulas/chemistry/high-school/jsjlgdynywt675qjv77qzhapj3jjpurbaf.png)
The new volume is unknown, but the temperature is increased to 85 degrees Celsius.
![(1.25 \ L)/(23 \textdegree C)=(V_2)/(85 \textdegree C)](https://img.qammunity.org/2022/formulas/chemistry/high-school/4wmh0lr36701xkglhoa7mppy4wwmlzve5h.png)
We are trying to solve for the new volume, V₂. It is being divided by 85 degrees Celsius. The inverse of division is multiplication, so we multiply both sides by 85°C.
![85 \textdegree C*(1.25 \ L)/(23 \textdegree C)=(V_2)/(85 \textdegree C)*85 \textdegree C](https://img.qammunity.org/2022/formulas/chemistry/high-school/dnolt47h17o1qfwkepa2djuex7vrgssypa.png)
![85 \textdegree C*(1.25 \ L)/(23 \textdegree C)= V_2](https://img.qammunity.org/2022/formulas/chemistry/high-school/7wdh8ypfrvff6d81573qc3qlzlqkt9zg1s.png)
The units of degrees Celsius cancel.
![85 *(1.25 \ L)/(23) = V_2](https://img.qammunity.org/2022/formulas/chemistry/high-school/q4g0fmjlr9dezq6odv3m26wzzmcftuhka8.png)
![4.61956522 \ L = V_2](https://img.qammunity.org/2022/formulas/chemistry/high-school/6s5v77lg07b7m4xngoh6dx6nz19t8kgghp.png)
The original measurements have at least 2 significant figures, so our answer must have 2. For the number we found, that is the tenth place.
The 1 in the hundredth place (in bold above) tells us to leave the 6 in the tenth place.
![4.6 \ L \approx V_2](https://img.qammunity.org/2022/formulas/chemistry/high-school/kbh5biui38z1i7wih6txlbumkc14txjzcb.png)
The new volume of the balloon is approximately 4.6 liters.