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

Answer:

a.  3.72 [atm]

Explanation:

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

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

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

$3.716957[atm]=P_2$

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