A balloon is filled with helium at a temperature of 23.0 degrees Celsius and a pressure of 775 mmHg so that its volume if 6.65 L. It then rises to an altitude where the temperature is 10.5 degrees Celsius and the pressure is 613 mmHg. What is the volume of the balloon at that altitude?
This question requires the use of the universal gas law PV=nRT. Since we are given P, V, and T at the beginning, we can solve for n, the number of moles of helium gas in the balloon. Since the pressure is given in millimeters of mercury (mmHg), we first need to convert that into atmospheres (remember 1 atm=760 mmHg):
775 mmHg * (1 atm/760 mmHg) = 1.02 atm
Now plug the different values into the gas equation:
n = (PV)/(RT)
n = (1.02 atm * 6.65 L) / (0.0821 L*atm/mole*K * 296.15 K) = 0.279 moles
Since the number of moles of helium gas does not change as the balloon rises, we can re-insert this value back into the equation with the new temperature and pressure values to get the new volume of the balloon. Remember to convert mmHg into atmospheres.
613 mmHg * (1 atm/760 mmHg) = 0.807 atm
V = (nRT)/P
V = (0.279 moles * 0.0821 L*atm/mole*K * 283.65 K) / 0.807 atm = 8.05 liters
The volume of the balloon at the new altitude is 8.05 liters.