Question

A 0.788 L balloon has a pressure of 1.20 atm. What is the pressure of the balloonwhen some naughty child squeezes it down to a volume of 0.650 L?

Answer 1

1.45 atm

Step-by-step explanation

Given:

The initial volume of the balloon, V₁ = 0.788 L

Initial pressure, P₁ = 1.20 atm

When some naughty child squeezes it down, the final volume, V₂ = 0.650 L

What to find:

The final pressure, P₂ when some naughty child squeezes the balloon down.

Step-by-step solution:

The final pressure, P₂ can be calculated using Boyle's law equation:

`\begin{gathered} P_1V_1=P_2V_2 \\ \\ \Rightarrow P_2=(P_1V_1)/(V_2) \end{gathered}`

Plug in the values of the given parameters into the equation above:

`P_2=\frac{1.20\text{ }atm*0.788\text{ }L}{0.650\text{ }L}=\frac{0.9456\text{ }atm}{0.650}=1.45\text{ }atm`

The pressure when some naughty child squeezes the balloon down = 1.45 atm.

The second option is the correct answer.