Question

A sealed balloon is filled with 1.00 L of helium at 19°C and 1.09 atm. The balloon rises to a point in the atmosphere where the pressure is 222 torr and the temperature is −28 °C. What is the change in the volume of the balloon as it ascends from 1.09 atm to a pressure of 222 torr?

Answer 1

Answer:

change in volume of ballon is  2.109 L

Explanation:

Given data:

initial temperature = 199 degree C = 292 K

pressure P1 = 1.09 atm

initial volume = 1.00 L

pressure P2 = 222 torr = 0.292 atm

temperature T2 = - 28 oC = 245 K

we know that$\frac{P1 V1}{T1} = \frac{P2 V2}{T2}$

$\frac{1.09 \times  1}{292} = \frac{0.292 \times  V2}{245}$

V2 = 3.109 L

change in volume of ballon = 3.109 - 1 = 2.109 L