Question

A weather balloon has a volume of 52.5 liters at a temperature of 295 K. The balloon is released and rises to an altitude where the temperature is 252 K. How does the temperature change described affect the gas particle motion?

Answer 1

So you have a balloon rising through the atmosphere. use the formula p1/v1=p2/v2 and add the variables into the equation, leaving 295/52.5=252/x. multiply 252 by 52.5 and divide that number by 295.
52.5*252=13230. divide by 295 =44.9 L

Answer 2

Answer:

Gas particle motion will drop as the temperature reaches 252 K

Explanation:

Initial temperature T1 = 295 K

Final temperature, T2 = 252 K

The kinetic energy of the gas particles is related to the temperature via the following equation:

$KE = \frac{3}{2} kT$

where k = Boltzmann constant

Based on the above equation we can write:

$\frac{KE(1)}{KE(2)} = \frac{T1}{T2} = \frac{295}{252} = 1.17$

KE(1) = 1.17KE(2)

This implies that the kinetic energy is lower at T2 and therefore the gas particles will slow down as the balloon rises.