Question

A weather balloon is filled to a volume of 6000 L while it is on the ground, at a pressure of 1 atm and a temperature of 273 K. The balloon is then released into the atmosphere. As the balloon rises, the pressure decreases to 0.3 atm and the temperature decreases to 240 K. What is the final volume of the balloon at this temperature and pressure? Round your answer to the nearest tenth.

Answer 1

We can use the combined gas law equation to solve this: P1V1/T1 = P2V2/T2

(1atm)(6000L) / (273K) = (0.3atm)V2 / (240K)

21.978 atmL/K = 0.00125atm/K V2

V2 = 17582.41758L

To the nearest tenth, the final volume is 17582.4L

Answer 2

Answer : The final volume of the balloon at this temperature and pressure is, 17582.4 L

Solution :

Using combined gas equation is,

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

where,

$P_1$ = initial pressure of gas = 1 atm

$P_2$ = final pressure of gas = 0.3 atm

$V_1$ = initial volume of gas = 6000 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 273 K

$T_2$ = final temperature of gas = 240 K

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{1atm\times 6000L}{273K}=\frac{0.3atm\times V_2}{240K}$

$V_2=17582.4L$

Therefore, the final volume of the balloon at this temperature and pressure is, 17582.4 L