Question

A sample of an ideal gas at 1.00 atm1.00 atm and a volume of 1.81 L1.81 L was placed in a weighted balloon and dropped into the ocean. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 40.0 atm,40.0 atm, what was the volume of the sample? Assume that the temperature was held constant.

Answer 1

`\boxed{\text{29.5 mL}}`

Step-by-step explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

`p_(1)V_(1) = p_(2)V_(2)`

Data:

`\begin{array}{rcrrcl}p_(1)& =& \text{1.00 atm}\qquad & V_(1) &= & \text{1.81L} \\p_(2)& =& \text{40.0}\qquad & V_(2) &= & ?\\\end{array}`

Calculations:

`\begin{array}{rcl}1.00 * 1.81 & =& 40.0V_(2)\\1.180 & = & 40.0V_(2)\\V_(2) & = &\textbf{0.0295 L = 29.5 mL}\\\end{array}\\\text{The volume of the gas is } \boxed{\textbf{29.5 mL}}`