Answer to Question #66730 in General Chemistry for Matt

Question

A weather balloon is partially filled with 0.285 m3 of helium
at 23.0°C and 757 Torr. Assuming the volume of gas can expand freely
in the balloon, what will be the volume when the balloon is at an
altitude where the temperature is –45°C and the pressure is 0.640 atm? Thanks!

Answer 1

For this task we use the ideal gas law:
PV=nRT, where
P is the pressure of the gas, Pa. Conversion of Torr to Pa: 1 Torr = 133.3 Pa; conversion of atm to Pa: 1 atm = 101325 Pa.
V is the volume of the gas, m3;
n is the number of moles of gas, mol;
R is the universal gas constant = 8.31 J mol−1 K−1;
T is the absolute temperature of the gas, K. Conversion of °C to K: K=°C+273.15.

First we find the number of moles of helium in the balloon:
n=PV/RT
Do the calculation:
n= (757*133.3) Pa * 0.285 m3 / 8.31 J mol−1 K−1 * (23.0+273.15) K = 11.686 moles.

Knowing the number of moles in the balloon and assuming that it remains unchanged we can find the volume at the new conditions (marked with index 1):

V1=nRT1/P1
Do the calculation:
V1= 11.686 mol * 8.31 J mol−1 K−1 * (-45+273.15) K / (0.640 * 101325) Pa = 0.342 m3


Answer:
The new volume of the balloon is 0.342 m3