A weather balloon is filled with 200L of helium at 27 degree Celsius and 0.950 atm. What would be the volume of the gas at -10 degrees and 0.125 atm?

Question

asked Dec 14, 2020 187k views

1 Answer

Julien Spronck · Answer 1 · 2020-12-19T21:13:26+0000

Answer: The volume when the pressure and temperature has changed is 1332.53 L

Step-by-step explanation:

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:

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

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=0.950atm\\V_1=200L\\T_1=27^oC=[273+27]K=300K\\P_2=0.125atm\\V_2=?L\\T_2=-10^oC=[273-10]K=263K$

Putting values in above equation, we get:

$(0.950atm* 200L)/(300K)=(0.125* V_2)/(263K)\\\\V_2=1332.53L$

Hence, the volume when the pressure and temperature has changed is 1332.53 L