A weather balloon contains 12.0 m3 of hydrogen gas when it is released from a location at which the temperature is 22°C and the pressure is 101...

This problem can be solved using the Combined Gas Law,

`(P_1V_1)/(T_1) = (P_2V_2)/(T_2)`

The Combined Gas Law gets its name because it combines Charles' Law which states that volume and temperature are directly proportional and and Boyle's Law which states that pressure and volume are inversely proportional. 

When solving a gas law problem, it's a good idea to organize and label the data don't you don't mix up values. Temperatures must be converted to Kelvins,...

This problem can be solved using the Combined Gas Law,

`(P_1V_1)/(T_1) = (P_2V_2)/(T_2)`

The Combined Gas Law gets its name because it combines Charles' Law which states that volume and temperature are directly proportional and and Boyle's Law which states that pressure and volume are inversely proportional. 

When solving a gas law problem, it's a good idea to organize and label the data don't you don't mix up values. Temperatures must be converted to Kelvins, by adding 273, because the Celsius temperture scale isn't proportional.

P1 = 101 kpa

V1 = 12.0 m^3

T1 = 22ºC = 295 K

P2 = 20 kpA

V2 = unknown

T2 = -30ºC = 242 K

Now rearrange the equation to solve for the unknown, T2, and plug in the other values:

`(V_2 = P_1V_1T_2)/(T_1P_2)`

V2 = [(101 kpa)(12.0 m^3)(242K)]/[(295 K)(20 kpa)] = 49.7 cm^3