Question

A very flexible helium-filled balloon is released from the ground into the air at 20 degrees celsius. The initial volume of the balloon is 5.00 L and the pressure is 760mmHg. The balloon ascends to an altitude of 20 km, where the pressure is 76.0 mmHg and the temperature is 50 degrees celsius. What is the new volume, V2, of the balloon in litres, assuming it doesn't break or leak?

Answer 1

To solve this problem, we can use the ideal gas law, which relates pressure, volume, and temperature of a gas. The ideal gas law is given by:

PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature of the gas in Kelvin.

We can assume that the mass of the helium in the balloon remains constant, so the number of moles of helium also remains constant. Therefore, we can write:

P1V1/T1 = P2V2/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature of the helium in the balloon, and P2, V2, and T2 are the final pressure, volume, and temperature of the helium at the higher altitude.

We are given:

P1 = 760 mmHg

V1 = 5.00 L

T1 = 20°C = 293 K

P2 = 76.0 mmHg

T2 = 50°C = 323 K

We can plug in these values and solve for V2:

P1V1/T1 = P2V2/T2

(760 mmHg)(5.00 L)/(293 K) = (76.0 mmHg)V2/(323 K)

V2 = (760 mmHg)(5.00 L)/(76.0 mmHg)(293 K/323 K)

V2 = 80.8 L

Therefore, the new volume of the balloon at the higher altitude is 80.8 L.