Question

In one city, a ballon with a volume of 6.0L is filled with air at standard pressure. The ballon is then taken to a second city at a much higher altitude. At this second city, atmospheric pressure is only 91 Kpa. If the temperature is the same in both places, what will be the new volume of the ballon?

Answer 1

Use the combined law.

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

Where the temperatures are the same, P1 = 101 kPa, P2 = 91 kPa, V1 = 6.0 L, and solve for V2.

`\begin{gathered} \frac{101\text{kPa}\cdot6L}{T}=\frac{91\text{kPa}\cdot V_2}{T} \\ V_2=(606)/(91)L \\ V_2=6.66L \end{gathered}`

Therefore, the new volume will be 6.66 L.