Question

A 10.5 L balloon contains helium gas at a pressure of 640 mmHg . What is the final pressure, in millimeters of mercury, of the helium gas at each of the following volumes if there is no change in temperature and amount of gas?Part A: 23.0 LPart B: 3.50 L

Answer 1

For part A: The final pressure of the helium gas = 292.2 mmHg

For part B: The final pressure of the helium gas = 1920 mmHg.

Step-by-step explanation

Part A:

The final pressure, in mmHg of the helium gas, can be calculated using Boyle's law formula:

`P_1V_1=P_2V_2`

Putting

`\begin{gathered} V_1=10.5\text{ }L \\ \\ P_1=640\text{ }mmHg \\ \\ V_2=23.0\text{ }L \\ \\ P_2=? \end{gathered}`

We have:

`\begin{gathered} 640mmHg*10.5L=P_2*23.0L \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }23.0L \\ \\ P_2=(640mmHg*10.5L)/(23.0L) \\ \\ P_2=292.2\text{ }mmHg \end{gathered}`

For part A: The final pressure of the helium gas is 292.2 mmHg.

Part B:

Similarly, the final pressure, in mmHg of the helium gas, can be calculated using Boyle's law formula:

`P_1V_1=P_2V_2`

Putting

`\begin{gathered} V_1=10.5\text{ }L \\ \\ P_1=640\text{ }mmHg \\ \\ V_2=3.50\text{ }L \\ \\ P_2=? \end{gathered}`

We have:

`\begin{gathered} 640mmHg*10.5L=P_2*3.50L \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }3.50L \\ \\ P_2=(640mmHg*10.5L)/(3.50L) \\ \\ P_2=1920\text{ }mmHg \end{gathered}`

For part B: The final pressure of the helium gas is 1920 mmHg.