Question

Initially, the balloon had 3.0 liters of gas at a pressure of 400 kPa and was at a temperature of 294 K. If the balloon is cooled to 277 K and its volume decreased to 1 L, what will the new pressure in the balloon be?

Answer 1

`(P1 V1)/(T1) = (P2 V2)/(T2)`

⇒ (400·3) / 294 = (P2·1) / 277

P2 = 1130.6 kPa