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.