A 6.77 L balloon contains helium at SATP at ground level. Predict the balloon’s temperature (in Kelvins) if it floats to an altitude where the balloon's volume shrinks to 3.05 L and the atmospheric pressure is 89.2 kPa.
To find the temperature of the balloon at the higher altitude, we can use the combined gas law equation:
(P1 x V1) / T1 = (P2 x V2) / T2
Where:
P1 = initial pressure at SATP = 101.3 kPa
V1 = initial volume = 6.77 L
T1 = initial temperature in Kelvins
P2 = final pressure at higher altitude = 89.2 kPa
V2 = final volume = 3.05 L
T2 = final temperature in Kelvins
Now we substitute the given values into the equation:
(101.3 kPa x 6.77 L) / T1 = (89.2 kPa x 3.05 L) / T2
648.74 / T1 = 271.86 / T2
Rearranging the equation to solve for T2:
T2 = (T1 x 271.86) / 648.74
We know that at SATP, the temperature is 273.15 K (0 degrees Celsius), so:
T2 = (273.15 K x 271.86) / 648.74
T2 = 114.87 K
Therefore, the temperature of the balloon at the higher altitude where its volume is 3.05 L and the atmospheric pressure is 89.2 kPa would be approximately 114.87 Kelvins.