Use Charles’ Law: V1/T1 = V2/T2. Given a mass of gas held at a constant pressure, changes in the gas’s temperature and volume are directly proportional. Since the volume of the gas in the balloon decreased, we should expect that the temperature had also decreased.
Here, V1 = 12.0 L, T1 = 357.15 K, and V2 = 5.0 L. To solve for T2, we rearrange the equation and compute:
T2 = V2T1/V1 = (5.0 L)(357.15 K)/(12.0 L) T2 = 148.8125 K - 273.15 = -124.34 °C
To two significant figures, the answer would be -120 °C.