Question

The pressure of a sample of helium is 1.556 atm in a 268.5 mL container. If the container is compressed to 112.4 mL without changing the temperature, what is the new pressure?

a. 3.72
b. 0.651
c. 1.94e4
d. 277

Answer 1

a. 3.72 [atm]

Step-by-step explanation:

For a gas at constant temperature, (with no change in number of molecules of the gas), we can apply Boyle's Law:
`P_1V_1=P_2V_2`

`(1.556[atm])(268.5[mL])=P_2(112.4[mL])`

`\frac{(1.556[atm])(268.5[mL\!\!\!\!\!\!\!\!{--}])}{112.4[mL \!\!\!\!\!\!\!\!{--}]}=\frac{P_2(112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})}{112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}`

`3.716957[atm]=P_2`

It seems like the answer should have 4 significant figures since all of the other quantities have 4 significant figures, but the closest answer choice of those provided is a. 3.72