Question

A sealed balloon is filled with 1.00 L of helium at 23°C and 1.00 atm. The balloon rises to a point in the atmosphere where the pressure is 284 torr and the temperature is -31°C. What is the change in the volume of the balloon as it ascends from 1.00 atm to a pressure of 284 torr?

Answer 1

Answer: The volume when the pressure and temperature has changed is 2.19 L

Explanation:

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

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

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:

Conversion factor used:  1 atm = 760 torr

$P_1=1.00atm\\V_1=1.00L\\T_1=23^oC=[23+273]=296K\\P_2=284torr=\frac{284}{760}=0.374atm\\V_2=?L\\T_2=-31^oC=[-31+273]K=242K$

Putting values in above equation, we get:

$\frac{1.00atm\times 1.00L}{296K}=\frac{0.374atm\times V_2}{242K}\\\\V_2=\frac{1.00atm\times 1.00L\times 242K}{0.374atm\times 296K}=2.19L$

Hence, the volume when the pressure and temperature has changed is 2.19 L