Question

A balloon is launched when the temperature is 15 oC and the pressure is 0.918 atm. Its volume is 11.1 L. It rises in the air until the temperature reaches -15 oC and the pressure is 0.0012 atm. What is its new volume?

Answer 1

Answer:

The new volume is 7,606.96 Liter.

Explanation:

The combined gas equation is,

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

where,

$P_1$ = initial pressure of gas in the balloon = 0.918 atm

$P_2$ = final pressure of gas in the balloon = 0.0012 atm

$V_1$ = initial volume of gas in the balloon = $11.1 L$

$V_2$ = final volume of gas in the balloon = ?

$T_1$ = initial temperature of gas in the balloon = $15^oC=273+15=288K$

$T_2$ = final temperature of gas in the balloon = $-15^oC=273-15=258K$

Now put all the given values in the above equation, we get:

$\frac{0.918 atm\times 11.1 L}{288 K}=\frac{0.0012 atm\times V_2}{258 K}$

$V_2=\frac{0.918 atm\times 11.1 L\times 258 K}{288 K\times 0.0012 atm}$

$V_2=7,606.97 L$

The new volume is 7,606.96 Liter.