A very flexible helium-filled balloon is released from the ground into the air at 20 degree C. The initial volume of the balloon is 5.00 L, and the pressure is 760 mmHg. The balloon ascends to an altitude of 20km, where the pressure is 76.0 mmHg and the temperature is -50 degree C. What is the new volume, V2, of the balloon in liters, assuming it doesn't break or leak?
(P1V1/T1) = (P2V2/T2)
T must be in kelvin.
To find the new volume, V2, of the balloon at an altitude of 20km, we can use the ideal gas law, which states that:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, we need to convert the given temperatures to Kelvin by adding 273.15:
Initial temperature, T1 = 20°C + 273.15 = 293.15 K
Final temperature, T2 = -50°C + 273.15 = 223.15 K
Next, we can use the ideal gas law to calculate the initial number of moles, n1, of the gas in the balloon at ground level. Assuming the helium behaves ideally, we can rearrange the ideal gas law equation to solve for n:
n = PV / RT
Substituting the given values:
P1 = 760 mmHg * (1 atm / 760 mmHg) = 1 atm
V1 = 5.00 L
T1 = 293.15 K
R = 0.0821 L.atm/mol.K
n1 = (1 atm * 5.00 L) / (0.0821 L.atm/mol.K * 293.15 K)
Calculate n1 using these values.
The next step is to use the ideal gas law again to find the final volume, V2, at an altitude of 20km. This time, we're given the pressure and temperature at that altitude, but the number of moles should remain constant because there is no mention of gas entering or leaving the balloon:
P2 = 76.0 mmHg * (1 atm / 760 mmHg) = 0.1 atm
T2 = 223.15 K
Rearranging the ideal gas law equation:
V2 = (n1 * R * T2) / P2
Substitute the known values and the calculated value of n1 to find V2.